Proving $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$

Question

For $a,b,c>0$, prove $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$

I've simplified the inequality by multiplying both sides with $(a+b+c).$

So the inequality became $\displaystyle a^2+b^2+c^2\geqslant \sum\limits_{cyc}a\frac{(b^2+c^2)}{b+c}$

From here I ve tried use some well known inequalities like Cauchy-Schwarz or QM-AM-GM-HM but as I m not very fluent with these yet I couldn't finish the proof.

Answer 1

We need to prove that: $$\sum_{cyc}\left(\frac{a^2+b^2+c^2}{a+b+c}-\frac{a^2+b^2}{a+b}\right)\geq0$$ or $$\sum_{cyc}\frac{c^2a+c^2b-a^2c-b^2c}{a+b}\geq0$$ or $$\sum_{cyc}\frac{ac(c-a)-bc(b-c)}{a+b}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{ab}{b+c}-\frac{ab}{c+a}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2ab}{(a+c)(b+c)}\geq0$$ and we are done!

Answer 2

You chose the right way.

The last inequality is equivalent to $$ab(a+b)(a-b)^2+bc(b+c)(b-c)^2+ca(c+a)(c-a)^2 \geqslant 0,$$ which is true.

Also$,$ you can re-write it in $uvw$ form$:$ $$9\, \left( u-v \right) ^{2}{w}^{3}+18\, \left( u-v \right) v{w}^{3}\\+27 \, \left( v-w \right) \left( 3\,{u}^{3}-4\,u{v}^{2}+{w}^{3} \right) v +27\, \left( 3\,{u}^{3}-4\,u{v}^{2}+{w}^{3} \right) vw \geqslant 0$$ And see it followes by AM-GM and Schur degree $3$.

Answer 3

We have $$\sum \frac{a^2+b^2}{a+b} \leqslant \frac{3(a^2+b^2+c^2)}{a+b+c},$$ equivalent to $$\sum \frac{[(a+b)^2-2ab](a+b+c)}{a+b} \leqslant 3(a^2+b^2+c^2),$$ or $$a^2+b^2+c^2 + 2abc\sum\frac{1}{a+b} \geqslant 2(ab+bc+ca).$$ Appy Cauchy-Schwarz Inequality, we have $$\sum \frac{1}{a+b} \geqslant \frac{9}{2(a+b+c)}.$$ Hence, we need to prove $$a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca),$$ or $$a^3+b^3+c^3+3abc \geqslant ab(a+b)+bc(b+c)+ca(c+a).$$ Which is Schur Inequality. We are done.