<s>[INST] Hard inequality for positive numbers The problem is to prove that for $a,b,c>0$ we have
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}$, and then to use Muirhead's inequalities to prove the remaining inequality - but unsuccessfully, so far...
 [/INST] I will use the following inequality:
(Vasile Cirtoaje) If $x,y,z$ are positive real numbers, then:
$$(x+y+z)^3 \geq \frac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
Setting $x=\frac{a^2}{b^2}, y = \frac{b^2}{c^2}, z = \frac{c^2}{a^2}$, we find
$$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^3\geq \frac{27}{4}\left(\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2}+1\right)$$
and using Cauchy-Schwarz
$$\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2} \geq \frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}$$
It remains to prove that:
$$\frac{27}{4}\left[\frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}+1\right]\geq \left[\frac{15}{4}-\frac{9abc}{4(a^3+b^3+c^3)}\right]^3$$
If we set
$$t=\frac{a^3+b^3+c^3}{abc}\geq 3$$
this becomes
$$\frac{27}{4}\left(\frac{t^2}{3}+3\right)\geq \left(\frac{15}{4}-\frac{9}{4t}\right)^3$$
Fully expanding this is equivalent with:
$$\frac{9}{64t^3}(t-3)(16t^4+48t^3-183t^2+126t-27)\geq 0$$
which is true as $t \geq 3$.
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