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.)
$a+x=b+y=c+z=1$
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]
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 $a=b$.
(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 $a=b$.
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