Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$

Question

Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$ Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$

My Solution.

We write the inequality as

$$\big[2(b+c)-a\big]^2 \big[4\left( {{\mkern 1mu} b + {\mkern 1mu} c} \right){a^2} + 5{\mkern 1mu} \left( {2{\mkern 1mu} c + b} \right)\left( {c + 2{\mkern 1mu} b} \right)a + 3{\mkern 1mu} \left( {b + c} \right)\left( {6{\mkern 1mu} {b^2} + 19{\mkern 1mu} bc + 6{\mkern 1mu} {c^2}} \right)\big]$$

$$+\big[a-2(b+c)\big]\left( {b}^{2}+3bc+{c}^{2} \right) \left( 36{b}^{2}+77bc+36{ c}^{2} \right)+2bc \left( b+c \right) \left( b-c \right) ^{2}\geq 0,$$

which is true.

You can see the text to check here. I'm hoping for alternative proof..

Thank you!

Answer 1

Let $c=\min\{a,b,c\}$, $b=c+u$ and $a-2(b+c)=v$.

Thus, $u\geq0$, $v\geq0$, $a=4c+2u+v$ and $$(ab+ac+bc)\prod_{cyc}(a+b)\left(5\sum_{cyc}\frac{a}{b+c}-\frac{a^2+b^2+c^2}{ab+ac+bc}-10\right)=$$ $$=745vc^4+2(2u^2+745uv+247v^2)c^3+(6u^3+1074u^2v+741uv^2+109v^3)c^2+$$ $$+(2u^4+329u^3v+355u^2v^2+109uv^3+8v^4)c+$$ $$+2uv(18u^3+27u^2v+13uv^2+2v^3)\geq0.$$

Answer 2

Suppose $a+b+c=1,$ let $p=a+b+c,\,q=ab+bc+ca,\,r=abc$ then $p \leqslant \frac{1}{3}.$ We write inequality as $$\frac{5(1-2q+3r)}{q-r} \geqslant \frac{1-2q}{q}+10,$$ or $$(1+23q)r+2q(2-9q) \geqslant 0. \quad (1)$$ If $0 < q \leqslant \frac{2}{9}$ then $(1)$ is true.

If $\frac{2}{9} < q \leqslant \frac{1}{4},$ from condition we get $$(a-2b-2c)(b-2c-2a)(c-2a-2b) \geqslant 0,$$ equivalent to $$r \geqslant \frac{18q-4}{27}.$$ Therefore, we need to prove $$(1+23q) \cdot \frac{18q-4}{27}+2q(2-9q) \geqslant 0,$$ or $$\frac{2(9q-2)(1-4q)}{27} \geqslant 0 \quad (\text{true}).$$ If $\frac{1}{4} < q \leqslant \frac 13,$ then $$\frac{a^2+b^2+c^2}{ab+bc+ca}+10 \leqslant 12.$$ We need to prove $$5\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \geqslant 12.$$ Let $t = \frac{a}{b+c} \geqslant 2,$ using the Cauchy-Schwarz inequality we have $$\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{(b+c)^2}{a(b+c)+2bc} \geqslant \frac{2}{2t+1}.$$ It's remain to prove that $$5t + \frac{10}{2t+1} \geqslant 12,$$ or $$\frac{(10t+1)(t-2)}{2t+1} \geqslant 0.$$ Which is true.