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I'm interested in the following problem :

Let $a,b,c>0$ be the variables and $u,v>0$ be constant then we have : $$\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$$

Rewrrting the inequality, we have :

$$\sum_{cyc}\frac{a}{a+b+c}\frac{1}{\frac{b}{a}u+\frac{b^2}{a^2}v}\geq \frac{1}{u+v}$$

As the function :

$$f(x)=\frac{1}{xu+x^2v}$$

Is convex (with the condition of positivity) we can apply Jensen's inequality and then we have:

$$\sum_{cyc}\frac{a}{a+b+c}f\Big(\frac{b}{a}\Big)\geq f\Big(\frac{a\frac{b}{a}+b\frac{c}{b}+c\frac{a}{c}}{a+b+c}\Big)=f(1)=\frac{1}{u+v}$$

Done !

My question is have you an alternative proof wich doesn't use Jensen's inequality ?

Thanks for sharing your time and knwoledge .

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By C-S $$\sum_{cyc}\frac{a^3}{uab+vb^2}=\sum_{cyc}\frac{a^4}{ua^2b+vb^2a}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(ua^2b+va^2c)}.$$ Thus, it's enough to prove that $$(u+v)\sum_{cyc}(a^4+2a^2b^2)\geq(a+b+c)\sum\limits_{cyc}(ua^2b+va^2c)$$ or $$\sum_{cyc}((u+v)a^4-ua^3b-va^3c)+(u+v)\sum_{cyc}(a^2b^2-a^2bc)\geq0,$$ which is true by Rearrangement and Muirhead.

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By Cauchy Schwarz / Titu's lemma,

$$ \sum \frac{a^4}{a^2bu+ab^2v} \geq \frac{(\sum a^2)^2}{ \sum a^2bu+ab^2v } $$

So it remains to show that

$$(u+v)(\sum a^2)^2 \geq \sum a \sum (a^2 bu + b^2av)$$

For this to be true, it likely has to be true for $u$, $v$ independently.
(At this step, we're hoping this would work. It could be that the inequality was already too strong, as would be in the case if you used $(bu + \frac{b^2}{a} v)$ as the denominator above.)

For $u$, it is obvious (Muirhead termwise, or AM-GM appropriately) that

$$\sum a^4 + 2 \sum a^2b^2 \geq \sum a^3b + a^2b^2 + a^2bc.$$

Same holds for $v$. So we are done.

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