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# Inequality 29(George Basdekis)

Problem:

If $\displaystyle x,y,z,r>0$ prove that

$\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq 4r\sum_{cyc}x^3y$.

We know that

$\displaystyle 4r\sum_{cyc}x^3y\leq \left(r\sum_{cyc}x^3y+1\right)^2=r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1$.

So, we only need to prove that

$\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1$,

or

$\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2+r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y$.

But from Cauchy-Schwarz Inequality we deduce that

$\displaystyle r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2$.

So ,we only need to prove now that

$\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2\geq 2r\sum_{cyc}x^3y$,

which is obviously true from Cirtoaje’s Inequality, Q.E.D.