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.

A collection of intriguing competition level problems for secondary school students.

Sunday, July 31, 2016

Wednesday, July 20, 2016

### 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:

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

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

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:

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

Tuesday, July 12, 2016

### 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)

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)

Sunday, July 10, 2016

### 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.

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.

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