Let $x,y,a,b$ be positive real numbers such that $x\not= y$, $x\not= 2y$, $y\not= 2x$, $a\not=3b$ and $\frac{2x-y}{2y-x}=\frac{a+3b}{a-3b}$. Prove that $\frac{x^2+y^2}{x^2-y^2}\ge 1$.
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Tags: algebra proposed, algebra
JohannesG
30.10.2010 21:46
${{\frac{2x-y}^{2y-x}}={\frac{a+3b}^{a-3b}}\Leftrightarrow{\frac{a}^{b}}={\frac{x+y}^{x-y}}$ which is positive, so $x>y$ $x^2+y^{2}\geq x^2-y^2$ $y^2\geq 0$ hope i'm right
sumanguha
01.11.2010 05:55
equivalently we can write $\frac{x+y}{x-y}=\frac{a}{b} $ or $\frac{x+y}{a}=\frac{x-y}{b}=k \neq 0 $ so then $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{(x+y)^{2}+(x-y)^{2}}{2(x+y)(x-y)}=\frac{a^{2}+b^{2}}{2ab} \geq 1 $ (last step AM-GM)
Arslan
05.03.2021 09:43
sumanguha wrote: equivalently we can write $\frac{x+y}{x-y}=\frac{a}{b} $ or $\frac{x+y}{a}=\frac{x-y}{b}=k \neq 0 $ so then $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{(x+y)^{2}+(x-y)^{2}}{2(x+y)(x-y)}=\frac{a^{2}+b^{2}}{2ab} \geq 1 $ (last step AM-GM) How do you know that (x-y) is strictly greater than 0?
huashiliao2020
08.05.2023 04:39
$\frac{x^2+y^2}{x^2-y^2}\geq 1$ <=> $2y^2\geq 0$