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Inequality 18(Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c,d$ be positive real numbers. Prove that

$\displaystyle \left(\frac{a}{a+b+c}\right)^{2}+\left(\frac{b}{b+c+d}\right)^{2}+\left(\frac{c}{c+d+a}\right)^{2}+\left(\frac{d}{d+a+b}\right)^{2}\geq\frac{4}{9}$.

Solution:

Let us use Holder’s inequality, that is

$\displaystyle \left[\sum_{cyc}\frac{a^2}{(a+b+c)^2}\right]\left[\sum_{cyc}a(a+b+c)\right]^{2}\geq \left(\sum_{cyc}a^{\frac{4}{3}}\right)^{3}$.

So, the current inequality reduces to:

$\displaystyle \sum_{cyc}\frac{a^2}{(a+b+c)^2}\geq \frac{\left(\sum_{cyc}a^{\frac{4}{3}}\right)^{3}}{\left[\sum_{cyc}a(a+b+c)\right]^{2}}=\frac{(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}+d^{\frac{4}{3}})^{3}}{\left[(a+c)^2+(b+d)^2+(a+c)(b+d)\right]^{2}}$.

Now, from the Power-Mean inequality we acquire that

$\displaystyle \frac{a^{\frac{4}{3}}+c^{\frac{4}{3}}}{2}\geq \left(\frac{a+c}{2}\right)^{\frac{4}{3}}$ and $\displaystyle \frac{b^{\frac{4}{3}}+d^{\frac{4}{3}}}{2}\geq \left(\frac{b+d}{2}\right)^{\frac{4}{3}}$.

And therefore we have that

$\displaystyle \sum_{cyc}\frac{a^2}{(a+b+c)^2}\geq \frac{1}{2}\cdot \frac{\left[(a+c)^{\frac{4}{3}}+(b+d)^{\frac{4}{3}}\right]^{3}}{\left[(a+c)^2+(b+d)^2+(a+c)(b+d)\right]^{2}}$.

Let us divide the last fraction by $\displaystyle (b+d)^2$. Then by setting $\displaystyle w=\left(\frac{a+c}{b+d}\right)^{\frac{1}{3}}$, the last inequality becomes

$\displaystyle \sum_{cyc}\frac{a^2}{(a+b+c)^2}\geq \frac{1}{2}\cdot \frac{(w^4+1)^3}{(w^6+w^3+1)^2}$.

So, we only need to prove that

$\displaystyle \frac{1}{2}\cdot \frac{(w^4+1)^3}{(w^6+w^3+1)^2}\geq \frac{4}{9}$,

or

$\displaystyle 9(w^4+1)^3\geq 8(w^6+w^3+1)^2$, or $\displaystyle 9\left(w^2+\frac{1}{w^2}\right)^{3}\geq 8\left(w^3+\frac{1}{w^3}+1\right)^{2}$.

Setting $\displaystyle t=w+\frac{1}{w}$ the above inequality becomes $\displaystyle 9(t^2-2)^2\geq 8(t^3-3t+1)^2$.

But the last inequality is equivalent to $\displaystyle (t-2)^2(t^4+4t^3+6t^2-8t-20)\geq 0$ which is obviously true since

$\displaystyle t^4+4t^3+6t^2-8t-20=t^4+4t^2(t-2)+4t(t-2)+10(t-2)\geq 0$

and

$\displaystyle t=w+\frac{1}{w}\geq 2$, Q.E.D.