# prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$

prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $$a,b,c$$

Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .

or as inequality is homogenous we take $$a+b+c=1$$.

or we have to prove (i am skipping the steps as it is just algebra) :

$$5(ab+bc+ca-abc)\ge 4(1+ab+bc+ca)(ab+bc+ca-3abc)$$

But i am not able to prove this by expanding.

How do i proceed?

Other methods are welcome!

Let $$p=a+b+c=1, \; q=ab+bc+ca, \; r=abc.$$ We need to prove $$5(ab+bc+ca-abc) \geqslant 4(1+ab+bc+ca)(ab+bc+ca-3abc),$$ equivalent to $$5(q-r) \geqslant 4(1+q)(q-3r),$$ or $$(12q+7)r \geqslant q(4q-1).$$ If $$4q-1 < 0,$$ then $$(12q+7)r > 0 > q(4q-1).$$ If $$4q-1 \geqslant 0,$$ from Schur inequality $$(a+b+c)^3+9abc \geqslant 4(a+b+c)(ab+bc+ca),$$ we get $$r \geqslant \frac{p(4q-p^2)}{9} = \frac{4q-1}{9}.$$ It's remain to prove that $$(12q+7) \cdot \frac{4q-1}{9} \geqslant q(4q-1),$$ or $$\frac{(3q+7)(4q-1)}{9} \geqslant 0.$$ Which is true. The proof is completed.

• That is very nice i forgot using $pqr$ Thank you!Will accept by today Oct 7, 2020 at 7:29
• Thank you, Albus Dumbledore =)) Oct 7, 2020 at 7:34

Proof by full expanding.

After your work we need to prove that: $$5\left(\left(\sum_{cyc}a\right)^3\sum_{cyc}ab-abc\left(\sum_{cyc}a\right)^2\right)\geq4\left(\left(\sum_{cyc}a\right)^2+\sum_{cyc}ab\right)\left(\sum_{cyc}a\sum_{cyc}ab-3abc\right)$$ or $$5\sum_{cyc}(a^2+2ab)\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\geq4\sum_{cyc}(a^2+3ab)\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{cyc}(a^2+2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq4\sum_{cyc}ab\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{cyc}(a^2-2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq0$$ or $$\sum_{cyc}(a^4b+a^4c+a^3b^2+a^3c^2+2a^2b^2c-2a^3b^2-2a^3c^2-4a^3bc-4a^2b^2c+$$ $$+10a^3bc+10a^2b^2c)\geq0$$ or $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2+6a^3bc+8a^2b^2c)\geq0,$$ which is true by Muirhed.

Of course, I could write a last line only, but I showed, how we can get it: three minutes of work!

• thank you very much , it really helps when you show how you expand otherwise i think it is only brute force.Again i don't think it is 3 minuteds work, for me coming up with this expansion would take me more time. I guess its because you are an inequality expert this is 3 minutes work! Oct 7, 2020 at 9:11
• @Albus Dumbledore You are welcome! Prove inequalities and you'll do it better. Oct 7, 2020 at 9:14

Also, $$uvw$$ helps here very well.

Indeed, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, after homogenization your last inequality it's $$f(w^3)\geq0,$$ where $$f$$ is a linear function,

which says that it's enough to prove it for the extreme value of $$w^3$$,

which by $$uvw$$ happens in the following cases.

1. $$w^3\rightarrow0^+$$.

Let $$c\rightarrow0^+$$.

Thus, we need to prove that:$$5ab\geq4(1+ab)ab$$ or $$ab\leq\frac{1}{4},$$ which is true by AM-GM: $$ab\leq\left(\frac{a+b}{2}\right)=\frac{1}{4}.$$

1. Two variables are equal.

Let $$b=a$$ and $$c=1-2a$$, where $$0

After this substitution we need to prove that: $$a(1-2a)(1-a)(18a^2-3a+1)\geq0,$$ which is obvious.