1
$\begingroup$

Find the maximum of the value $a$ such foy any real postive numbers $x,y,z$ have $$\sqrt{\dfrac{x}{ax+y+z}}+\sqrt{\dfrac{y}{ay+z+x}}+\sqrt{\dfrac{z}{az+x+y}}\le3\sqrt{\dfrac{1}{2+a}}$$

I conjecture $a>0?$

$\endgroup$
0
$\begingroup$

It's wrong for $a\rightarrow0^+$. Try $x\rightarrow+\infty.$

By the way, just by Jensen your inequality is true for all $a\geq\frac{3}{4}.$

Also, by Vasc's LCF Theorem it's enough to prove your inequality for $z=y$

and since our inequality is homogeneous, we can assume $y=z=1$, which gives a minimal value of $a$, for which our inequality is true: $a=\frac{1}{3}$

For $a=\frac{1}{3}$ we get the following interesting inequality.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\sqrt{\frac{a}{a+3b+3c}}+\sqrt{\frac{b}{b+3c+3a}}+\sqrt{\frac{c}{c+3a+3b}}\leq\frac{3}{\sqrt7}$$ The equality occurs also for $a=b=1$ and $c=8$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.