<s>[INST] Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality 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 .
 [/INST] 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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