Solve x³+3xy²=-49 and x²-8xy+y²=8y-17x

Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$

My solution:

From $x^2-8xy+y^2=8y-17x$, we do some algebraic manipulation to get

\[ (y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1) \]

\[ x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2) \]

From $x^3+3xy^2=-49$, in trying to relate the $y$ term to $y-4x$, we get

\[ x^3+3xy^2=-49\\x^3+3x(y-4x+4x)^2=-49\\x^3+3x[(y-4x)^2+8x(y-4x)+16x^2]=-49\\x^3+3x(y-4x)^2+24x^2(y-4x)+48x^2=-49\\49(x^3+1)+3x[(y-4x)^2+8x(y-4x)]=0\\49(x^3+1)+3x(8y-17x+15x^2+8xy-32x^2)=0\text{   from (1)}\\49(x+1)(x^2-x+1)+3x[8y(x+1)-17x(x+1)]=0\\(x+1)[49(x^2-x+1)+3x(8y-17x)]=0\\(x+1)[49x^2-49x+49+24xy-51x^2]=0\\(x+1)[-2x^2-49x+49+3(8xy)]=0\\(x+1)[-2x^2-49x+49+3x^2+3y^2-24y+51x]=0\text{   from (2)}\\(x+1)(x^2+2x+1+3y^2-24y+48)=0\\(x+1)[(x+1)^2+3(y-4)^2]=0 \]

This is possible if and only if $x=-1$.

When $x=-1$, we get

\[ -1-3y^2=-49\\y^2=16\\y=\pm 4 \]

 

Find the sum \(a+b+c+d\).

Consider the following:

[MATH]f(x)=x^2-rax-sb[/MATH]

which has zeros \(c,d\).

[MATH]g(x)=x^2-rcx-sd[/MATH]

which has zeros \(a,b\).

Given \(f(x)\ne g(x)\), find the sum \(a+b+c+d\) in terms of \(r\) and \(s\) only.

Mark's solution:

Consider the following:

[MATH]f(x)=x^2-rax-sb[/MATH]

which has zeros \(c,d\).

[MATH]g(x)=x^2-rcx-sd[/MATH]

which has zeros \(a,b\).

Given \(f(x)\ne g(x)\), find the sum \(a+b+c+d\) in terms of \(r\) and \(s\) only.

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

Minimize [MATH]\prod_{i=1}^{2017}x_i[/MATH].

If $x_1,x_2,...,x_{2017} \in\Bbb{R}^+$ and [MATH]\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+...+\dfrac{1}{1+x_{2017}} = 1[/MATH], find the minimal possible value of [MATH]\prod_{i=1}^{2017}x_i[/MATH].

Solution:

By cyclic symmetry, we know the critical value is at the point:

[MATH]\left(x_1,\cdots,x_{2017}\right)=(2016,\cdots,2016)[/MATH]

And the objection function at that point is:

[MATH]f(2016,\cdots,2016)=2016^{2017}[/MATH]

Now, looking at another point on the constraint:

[MATH]\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)[/MATH]

We find the objective function at that point is:

[MATH]f\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)=\frac{2^{2016}2016^{2017}}{2017}>2016^{2017}[/MATH]

And so we conclude:

[MATH]f_{\min}=2016^{2017}[/MATH]

 

You can find ❤️ in ...... Math!

Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition: $a+x=b+y=c+z=1$ Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH].

Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition:

$a+x=b+y=c+z=1$

Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH].

My solution:

Rewrite the intended LHS of the inequality strictly in terms of a, b and c, we have:

$x=1-a,\,y=1-b,\,z=1-c$

[MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)[/MATH]

[MATH]=\left(abc+(1-a)(1-b)(1-c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(abc+1+ab+bc+ca-(a+b+c)-abc\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(1+ab+bc+ca-(a+b+c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(1-a(1-b)-b(1-c)-c(1-a)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}-1-a(1-b)\left(\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]\,\,\,\,\,\,-1-b(1-c)\left(\frac{1}{a(1-b)}+\frac{1}{c(1-a)}\right)-1-c(1-a)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}\right)[/MATH]

[MATH]=\frac{1-b(1-c)-c(1-a)}{a(1-b)}+\frac{1-a(1-b)-c(1-a)}{b(1-c)}+\frac{1-a(1-b)-b(1-c)}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-b+bc-c+ca}{a(1-b)}+\frac{1-a+ab-c+ac}{b(1-c)}+\frac{1-a+ab-b+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-b-c(1-b)+ca}{a(1-b)}+\frac{1-c-a(1-c)+ab}{b(1-c)}+\frac{1-a-b(1-a)+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{(1-b)(1-c)+ca}{a(1-b)}+\frac{(1-a)(1-c)+ab}{b(1-c)}+\frac{(1-b)(1-a)+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-c}{a}+\frac{c}{1-b}+\frac{1-a}{b}+\frac{a}{1-c}+\frac{1-b}{c}+\frac{b}{1-a}-3[/MATH]

[MATH]\ge 6\sqrt[6]{\frac{1-c}{a}\cdot \frac{c}{1-b}\cdot\frac{1-a}{b}\cdot\frac{a}{1-c}\cdot\frac{1-b}{c}\cdot\frac{b}{1-a}}-3[/MATH] (By the AM-GM inequality, with $1-a,\,1-b,\,1-c$ are all positive)

[MATH]= 6-3[/MATH]

[MATH]= 3[/MATH] (Q.E.D.)

Evaluate [MATH]\small\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.

Evaluate [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.

My solution:

[MATH]\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)[/MATH]

[MATH]=\frac{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}{\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}[/MATH]

[MATH]=\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}[/MATH]

By the Cauchy-Schwarz inequality, we have:

[MATH]\begin{align*}\sqrt{2}+\sqrt{3}+\sqrt{6}&<\sqrt{1+1+1}\sqrt{2+3+6}\\&=\sqrt{33}\end{align*}[/MATH]

Hence [MATH]\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}>\frac{23}{\sqrt{33}}[/MATH].

From $528\lt 529$ we get, after taking the square root on both sides and rearranging:

$4\lt \dfrac{23}{\sqrt{33}}$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\gt \dfrac{23}{\sqrt{33}}\gt 4$

On the other hand,

From $50\gt 49$, we get:

$\sqrt{2}\gt \dfrac{7}{5}$

From $12\gt 9$, we get:

$\sqrt{3}\gt \dfrac{3}{2}$

From $6\gt 4$, we get:

$\sqrt{6}\gt 2$

Adding them up gives:

$\sqrt{2}+\sqrt{3}+\sqrt{6}\gt 4.9$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\lt \dfrac{23}{4.9}=4.69$.

We can conclude by now that [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor=4.[/MATH]

If $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, prove [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH].

Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then
(a) [MATH]\frac{c}{\sqrt{ab}}\ge \sqrt{2}[/MATH]
(b) [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH].

My solution:

(a)

$\begin{align*}c^2&=a²+b²\\&≥2ab \,\,\,\text{(by the AM-GM inequality)}\end{align*}$

$\dfrac{c^2}{ab}\ge 2$
Taking the square root on both sides completes the proof, i.e. $\dfrac{c}{\sqrt{ab}}\ge \sqrt{2}$, equality occurs when $a=b$.

(b)

We have $c^2−a^2=b^2 \implies c−a=\dfrac{b^2}{c+a}$.

By the similar token, we also have $c−b=\dfrac{a^2}{c+b}$, if we're going to replace these two into the original LHS of the inequality, we get:

[MATH]\begin{align*}\frac{(c − a)(c − b)}{(c + a)(c + b)}&=\frac{(b^2)(a^2)}{(c + a)^2(c + b)^2}\\&=\left(\frac{ab}{(c + a)(c + b)}\right)^2\\&=\left(\frac{ab}{c^2+c(a+b)+ab}\right)^2\\&=\left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(a+b)}{ab}+1}\right)^2\\& \le \left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(2\sqrt{ab})}{ab}+1}\right)^2\,\,\text{(by the AM-GM inequality)}\\& = \left(\frac{1}{\left(\dfrac{c}{\sqrt{ab}}\right)^2+\dfrac{2c}{\sqrt{ab}}+1}\right)^2\\& \le \left(\frac{1}{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\right)^2\\&= \left(3-2\sqrt{2}\right)^2\\&= 17-12\sqrt{2}\,\,\,\text{Q.E.D.}\end{align*}[/MATH]

Equality occurs when $a=b$.

Solve for real solution of the system below: [MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

Solve for real solution of the system below:

[MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

My solution:

First, observe that the LHS of the equality must yield an integer, this tells us the fractional part of $x$ must be a zero, so this turns the whole equation as:

[MATH]x^3+x^2+x=-1\\x^3+x^2+x+1=0\\(x+1)(x^2+1)=0[/MATH]

This implies $x=-1$ is the only real solution to the system.

Solve for real solutions for the system: $x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

Solve for real solutions for the system:

$x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

My solution:

Let $x,\,y$ and $z$ be the real roots for a cubic polynomial.

From the relation $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we have:

$(-a)^2=a^2+2(xy+yz+zx)$

$a^2=a^2+2(xy+yz+zx)\implies xy+yz+zx=0$

From the relation $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$, we have:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

$-a^3-3xyz=(-a)(a^2-0)$

$-a^3-3xyz=-a^3\implies xyz=0$

We can now form the cubic polynomial in $t$ where its roots are $x,\,y$ and $z$:

$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0$

$t^3-(-a)t^2+(0)t-0=0$

$t^2(t+a)=0$

Obviously $t=0$ is the repeated root and the other root is $t=-a$.

Therefore we get the solution:

$(x,\,y,\,z)=(0,\,0,\,-a),\,(0,\,-a,\,0),\,(-a,\,0,\,0)$

Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

My solution:

[MATH](ab+bc+ca)(a^2+b^2+c^2)^2[/MATH]

[MATH]\ge \frac{(ab+bc+ca)(a^2+b^2+c^2)(a+b+c)^2}{3}[/MATH] since $3(a^2+b^2+c^2)\ge (a+b+c)^2$

[MATH]= \frac{(ab+bc+ca)(a^2+b^2+c^2)}{3}[/MATH] since $a+b+c=1$

[MATH]= \frac{a^3b+b^3c+ac^3+a^3c+ab^3+bc^3+a^2bc+ab^2c+abc^2}{3}[/MATH]

[MATH]= \frac{\left(\frac{a^2}{\frac{1}{ab}}+\frac{b^2}{\frac{1}{bc}}+\frac{c^2}{\frac{1}{ca}}\right)+\left(\frac{a^2}{\frac{1}{ac}}+\frac{b^2}{\frac{1}{ab}}+\frac{c^2}{\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH]

[MATH]\ge \frac{\left(\frac{(a+b+c)^2}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\right)+\left(\frac{(a+b+c)^2}{\frac{1}{ac}+\frac{1}{ab}+\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH] (By the Titu's Lemma)

[MATH]= \frac{\left(\frac{1}{\frac{a+b+c}{abc}}\right)+\left(\frac{1}{\frac{a+b+c}{abc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH]

(since [MATH]a+b+c=1[/MATH] and [MATH]\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{c}{abc}+\frac{a}{abc}+\frac{b}{cab}=\frac{a+b+c}{abc})[/MATH]

[MATH]= \frac{\left(abc+abc\right)+abc(a+b+c)}{3}[/MATH]

[MATH]= \frac{3abc}{3}[/MATH]

[MATH]= abc[/MATH] (Q.E.D)

Solve for real solution for $(1+x^2)(1+x^3)(1+x^5)=8x^5$.

Solve for real solution for $(1+x^2)(1+x^3)(1+x^5)=8x^5$.

My solution:

For $x\lt 0$, we have a positive left hand side value and a negative right hand side value. So $x$ can never be a negative value.

For $x\gt 1$, we have:

$1+x^2\gt 2x,\,(1+x^3)(1+x^5)=1+x^3+x^5+x^8\gt 4x^4$ so $(1+x^2)(1+x^3)(1+x^5)\gt 8x^5$, which really is $8x^5\gt 8x^5$, which leads to a contradiction.

For $0\le x \le 1$:

$f(x)=(1+x^2)(1+x^3)(1+x^5)$ has its first derivative of $f'(x)\gt 0$ and so $f$ is an increasing function and so does $f(x)=8x^5$.

That means they can intersect at most once, and by inspection, it is not hard to see that $x=1$ is the only real solution to the system.

Find the real solution(s) to the system $\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

Find the real solution(s) to the system

$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

My solution:

Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

My solution:

Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that

[MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

My solution:

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

Hello readers!

In my previous blog post, I asked the readers to spot the factual mistake(s) that I might have or might not have made in the solution (of mine) to one delicious inequality problem.

Today, I am going to discuss with you the mistake that I intentionally made.

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

Hello all!

Today I'm going to post something that is going to be very different than my style in my previous blog posts, as today I wanted to train students to spot the factual mistake(s) that I might have or might not have made in the following solution (of mine) to today's delicious inequality problem.

Prove the following inequality holds: $\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

My solution:

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

Prove that $x^2 + y^2+ z^2\le xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

Prove that $x^2 + y^2+ z^2\le  xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

For all $x,\,y,\, z\in [0,1]$, we know $x^2 + y^2+ z^2\le x+y+z$.