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# Inequality 14(Unknown Author)

Problem:

Let $\displaystyle a,b,c>0$ with $\displaystyle a+b+c=1$. Show that

$\displaystyle \frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq 2$.

Solution:

Subtract each side with $\displaystyle -1$. Then, from hypothesis we have that

$\displaystyle \frac{a^2+b}{b+c}-a+\frac{b^2+c}{c+a}-b+\frac{c^2+a}{a+b}-c\geq 1$.

Or

\displaystyle \begin{aligned}\frac{a^2+b}{b+c}-a+\frac{b^2+c}{c+a}-b+\frac{c^2+a}{a+b}-c=\frac{a(a-c)+b(b+c)}{b+c}&+\frac{b(b-a)+c(c+a)}{c+a}\\&+\frac{c(c-b)+a(a+b)}{a+b}\geq 1\end{aligned}.

From the last relation we have that

$\displaystyle \sum_{cyc}\frac{a(a-c)}{b+c}+\sum_{cyc}a\geq 1\Longrightarrow \sum_{cyc}\frac{a(a-c)}{b+c}\geq 0$.

Assume without loss of generality that $\displaystyle 0.

Then we have that

$\displaystyle \frac{b(b-a)}{c+a}+\frac{c(c-b)}{a+b}\geq \frac{a(c-a)}{b+c}$.

Doing some manipulations on both sides we acquire that

$\displaystyle b^4+b^3c-a^2b^2-a^2bc+c^4+c^3a-b^2c^2-ab^2c\geq a^2c^2+abc^2-a^4-a^3b$.

Or,

$\displaystyle \sum_{cyc}a^4+\sum_{cyc}a^3b\geq \sum_{cyc}a^2b^2+abc\sum_{cyc}a$,

which holds.

Indeed, from the AM-GM inequality we have that

$\displaystyle \sum_{cyc}a^4\geq \sum_{cyc}a^2b^2$

and from Cauchy-Schwartz inequality

$\displaystyle \sum_{cyc}\frac{a^2}{c}\geq \sum_{cyc}a\Longrightarrow \sum_{cyc}a^3b\geq abc\sum_{cyc}a$.

Adding up these $\displaystyle 2$ inequalities we get the desired result, Q.E.D.