Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$

Question

The following is an inequality which I have trouble solving: $$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$ ($a, b, c>0$)

I tried multiplying LHS and RHS by 2 and then use $${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$ and move to LHS but did not succeed in solving by this means.

Please provide a solution.

Answer 1

Another way. $$\sum_{cyc}\frac{a^2+bc}{b+c}-a-b-c=\sum_{cyc}\left(\frac{a^2+bc}{b+c}-\frac{b+c}{2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2-c^2}{b+c}=\frac{1}{2}\sum_{cyc}\frac{a^2-b^2-(c^2-a^2)}{b+c}=$$ $$=\frac{1}{2}\sum_{cyc}(a^2-b^2)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a+b)}{(a+c)(b+c)}\geq0.$$

Answer 2

Because $$\sum_{cyc}\left(\frac{a^2+bc}{b+c}-a\right)=\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{\prod\limits_{cyc}(a+b)}=\frac{\sum\limits_{cyc}(a^2-b^2)^2}{2\prod\limits_{cyc}(a+b)}\geq0.$$

Answer 3

Another way.

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\left(\frac{a^2+bc}{b+c}-a\right)=\sum_{cyc}\frac{(a-b)(a-c)}{b+c}\geq\frac{(a-b)(a-c)}{b+c}+\frac{(b-a)(b-c)}{a+c}=$$ $$=(a-b)\left(\frac{a-c}{b+c}-\frac{b-c}{a+c}\right)\geq0$$ because $a-b\geq0,$ $a-c\geq b-c$ and $\frac{1}{b+c}\geq\frac{1}{a+c}.$

Answer 4

Another way.

By C-S $$\sum_{cyc}\frac{a^2+bc}{b+c}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+c)}=$$ $$=\frac{\sum\limits_{cyc}(a^4+2a^2b^2+a^2b^2+2a^2bc+2a^3b+2a^3c+2a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}=$$ $$=a+b+c+\frac{\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}-\sum_{cyc}a=$$ $$=a+b+c+\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{2\sum\limits_{cyc}(a^2b+a^2c)}=a+b+c+\frac{\sum\limits_{cyc}(a^2-b^2)^2}{4\sum\limits_{cyc}(a^2b+a^2c)}\geq a+b+c.$$

Answer 5

Another way.

Since $(x+y+z)^2\geq3(xy+xz+yz),$ by C-S we obtain: $$\sum_{cyc}\frac{a^2+bc}{b+c}=\sqrt{\left(\sum_{cyc}\frac{a^2+bc}{b+c}\right)^2}\geq\sqrt{3\sum_{cyc}\frac{(a^2+bc)(b^2+ac)}{(b+c)(a+c)}}=$$ $$=\sqrt{3(a^2+b^2+c^2)}\geq a+b+c.$$ The following inequality is also true.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq3\left(\frac{a^{\frac{20}{9}}+b^{\frac{20}{9}}+c^{\frac{20}{9}}}{3}\right)^{\frac{9}{20}}.$$