[MATH]\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}[/MATH] for all positive reals $a$ and $b$.
My solution:
Step 1:
Squaring both sides of the inequality:
[MATH]\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}[/MATH]
[MATH]\left(\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\right)^2\le (\sqrt{2})^2[/MATH]
[MATH]\frac{a^2+b^2}{(a+b)^2}+2\left(\frac{\sqrt{a^2+b^2}}{a+b}\cdot \sqrt{\frac{ab}{a^2+b^2}}\right)+\frac{ab}{a^2+b^2}\le 2[/MATH]
[MATH]\frac{a^2+b^2}{(a+b)^2}+\frac{2\sqrt{ab}}{a+b}+\frac{ab}{a^2+b^2}\le 2[/MATH]
[MATH]\frac{a^2+b^2}{(a+b)^2}+\frac{ab}{a^2+b^2}+\frac{2\sqrt{ab}}{a+b}\le 2[/MATH](*)
Up to this point, if we can prove the above inequality (*) as correct, then we're done.
Step 2:
Observe that the last term on the LHS of the inequality (*) above is less than or equal to 1, since:
[MATH]a+b\ge 2\sqrt{ab}[/MATH]
[MATH]\frac{2\sqrt{ab}}{a+b}\le 1[/MATH]
It remains to show [MATH]\frac{a^2+b^2}{(a+b)^2}+\frac{ab}{a^2+b^2}-1\le 0[/MATH](**).
Step 3:
If we distribute the $-1$ as two negative one half and give it to the two terms on the LHS of (**), we see that we have:
[MATH]\frac{a^2+b^2}{(a+b)^2}+\frac{ab}{a^2+b^2}-1[/MATH]
[MATH]=\frac{a^2+b^2}{(a+b)^2}-\frac{1}{2}+\frac{ab}{a^2+b^2}-\frac{1}{2}[/MATH]
[MATH]=\frac{2a^2+2b^2-(a+b)^2}{2(a+b)^2}+\frac{2ab-a^2-b^2}{2(a^2+b^2)}[/MATH]
[MATH]=\frac{2a^2+2b^2-a^2-2ab-b^2}{2(a+b)^2}-\frac{a^2-2ab+b^2}{2(a^2+b^2)}[/MATH]
[MATH]=\frac{a^2-2ab+b^2}{2(a+b)^2}-\frac{a^2-2ab+b^2}{2(a^2+b^2)}[/MATH]
[MATH]=\frac{(a-b)^2}{2(a+b)^2}-\frac{(a-b)^2}{2(a^2+b^2)}[/MATH]
[MATH]=(a-b)^2\left(\frac{1}{2(a+b)^2}-\frac{1}{2(a^2+b^2)}\right)[/MATH]
[MATH]=(a-b)^2\left(\frac{a^2+b^2-(a+b)^2}{2(a+b)^2(a^2+b^2)}\right)[/MATH]
[MATH]=(a-b)^2\left(-\frac{2ab}{2(a+b)^2(a^2+b^2)}\right)[/MATH]
[MATH]\le 0[/MATH] since both $a$ and $b$ are positive real numbers and the inequality follows.
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