lucia2016 wrote: Let $x,y>0$ be real numbers.Prove that: $$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2}$$I tried CBS, but it doesn't work... Can you give an idea, please? See also here http://www.artofproblemsolving.com/community/c6h1122010p5167761

huynguyen wrote: $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2+y^2}\geq\frac{10}{(x+y)^2}$, which is just AM-GM. I don't realize how to demonstrate it with AM-GM...

lucia2016 wrote: Let $x,y>0$ be real numbers.Prove that: $$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2}$$I tried CBS, but it doesn't work... Can you give an idea, please? $\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{1}{x^2+y^2} +\frac{1}{2xy}+\frac{1}{2xy}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{25}{x^2+y^2+8xy}\ge \frac{10}{(x+y)^2}$

Wow, thank you sqing.

We can also solve using cauchy schwarz inequality on: $(\frac{1}{\sqrt{x^{2}+y^{2}}},\frac{1}{x},\frac{1}{y})$ and $(\sqrt{x^2+y^2},2x,2y)$ Giving: $(\frac{1}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2})(x^2+y^2+4x^2+4y^2)\geq (1+2+2)^2$ $\frac{1}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{25}{5(x^2+y^2)}$ $\frac{1}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{5}{x^2+y^2}$ $\frac{1}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{5}{\frac{(x+y)^2+(x-y)^2}{2}}$ $\frac{1}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{10}{(x+y)^2}$

It's wrong. I've try that before, too. $$\frac{10}{(x+y)^2+(x-y)^2}\le\frac{10}{(x+y)^2}$$

lucia2016 wrote: huynguyen wrote: $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2+y^2}\geq\frac{10}{(x+y)^2}$, which is just AM-GM. I don't realize how to demonstrate it with AM-GM... $\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{x^2+y^2} +\frac 34 .\left (\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac 14.\left (\frac{1}{x^2}+\frac{1}{y^2}\right)$ AM-GM $\frac 34.\left (\frac{1}{x^2}+\frac 1{y^2}\right)\geq \frac 34. \frac 8{(x+y)^2}=\frac 6{(x+y)^2}$ $\frac 14.\left (\frac{1}{x^2}+\frac 1{y^2}\right )\geq \frac 1{2xy}$ $\frac 1{2xy}+ \frac 1{x^2+y^2}\geq \frac {(1+1)^2}{2xy+x^2+y^2}=\frac 4{(x+y)^2}$ Hence , we are done.

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2+y^2} \ge \frac{2}{xy}+\frac{1}{x^2+y^2} =\frac{3}{2xy}+\frac{1}{2xy}+\frac{1}{x^2+y^2} $ $\ge \frac{3}{2xy}+\frac{4}{(x+y)^2} \ge \frac{10}{(x+y)^2}$ It remains to prove that $\frac{3}{2xy} \ge \frac{6}{(x+y)^2} \iff$ $x^2+y^2 \ge 2xy$ wich is true.

Nice solution ali3985!

lucia2016 wrote: Let $x,y>0$ be real numbers.Prove that: $$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2}$$I tried CBS, but it doesn't work... Can you give an idea, please? The following inequality is also true

$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2} \iff$ $\frac{1}{x^2+y^2} - \frac{2}{(x+y)^2} +(\frac{1}{x}-\frac{1}{y})^2 \ge \frac{8}{(x+y)^2}-\frac{2}{xy} \iff$ $ \frac{-(x-y)^2}{(x^2+y^2)(x+y)^2} +(\frac{1}{x}-\frac{1}{y})^2 \ge \frac{-2(x-y)^2}{xy(x+y)^2} \iff$ $ \frac{(x-y)^2( x^2+y^2+(x-y)^2)}{(x^2+y^2)(x+y)^2} +(\frac{1}{x}-\frac{1}{y})^2 \ge 0 $

Quote: The following inequality is also true We have by AM-GM $\frac{1}{x^2+y^2}+\frac{1}{2xy} \ge\frac{4}{(x+y)^2} \iff$ $\frac{8}{x^2+y^2}+\frac{4}{xy} \ge\frac{32}{(x+y)^2} \dots (1)$ and $\frac{4}{x^2+y^2}+\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{x^2+y^2}+\frac{x^2+y^2}{x^2y^2} \ge\frac{4}{xy} \dots (2) $ Adding $(1)$ and $(2)$ we get the desired inequality.

lucia2016 wrote: Let $x,y>0$ be real numbers.Prove that: $$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2}$$I tried CBS, but it doesn't work... Can you give an idea, please? We have \[\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2} - \frac{10}{(x+y)^2} = \frac{(x^4+4x^3y+x^2y^2+4xy^3+y^4)(x-y)^2}{x^2y^2(x^2+y^2)(x+y)^2} \geqslant 0.\] luofangxiang wrote: The following inequality is also true We have \[\frac{1}{x^2}+\frac{1}{y^2}+\frac{12}{x^2+y^2}-\frac{32}{(x+y)^2}=\frac{(x^2+6xy+y^2)(x-y)^4}{x^2y^2(x^2+y^2)(x+y)^2} \geqslant 0.\]

beautiful，Nguyenhuyen_AG

\[\frac{1}{x^2}+\frac{1}{y^2} \geq \frac{2}{xy}\]\[LHS \geq \frac{1}{x^2+y^2}+\frac{2}{xy}\]\[\frac{1}{x^2+y^2}+\frac{2}{xy} \geq \frac{10}{(x+y)^2}\]\[(x+y)^2(2x^2+2y^2+xy) \geq 10xy(x^2+y^2)\]\[5(x^3y+xy^3)+2x^4+2y^4+6x^2y^2 \geq 10x^3y+10xy^3\]\[2x^4+2y^4+6x^2y^2 \geq 5x^3y+5xy^3\]Inequality is homogeneous. Let $y=1$. \[2x^4+6x^2+2 \geq 5x^3+5x\]\[(x-1)^2(2x^2-x+2) \geq 0\]The last inequality is obviously true.

lucia2016 wrote: Let $x,y>0$ be real numbers.Prove that: $$\frac{1}{x^2+y^2} +\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{10}{(x+y)^2}$$ Prove that for any positive real $x$ and $y$, holds the inequality $$\frac{1}{(x+y)^2}+\frac{1}{x^2}+\frac{1}{y^2} \ge \frac{9}{4xy}$$