Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

Question

Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds:

$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$

What I have tried is firstly using the inequality:

$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, for any $x, y$ and any $a,b > 0$.

Using this inequality we obtain $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} \geq \frac{a + b + c}{2}$, and then we have: $$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq \frac{a + b + c}{2} + \frac{36}{a + b + c} = \frac{(a + b + c)^2 + 72)}{2(a + b + c)} $$. Using $ab + bc + ca = 1$, we would then have to prove that: $$a^2 + b^2 + c^2 + 74 - 40(a + b + c) \geq 0 $$ and then I tried replacing in this inequality $c = \frac{1 - ab}{a + b}$, but I didn't get anything nice.

I also tried rewriting the lhs: $$\frac{a^2}{b + c} = \frac{a^2(ab + bc + ca)}{b + c} = a^3 + \frac{a^2bc}{b + c}$$ And this would result in: $a^3 + b^3 + c^3 + abc(\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}) + \frac{36}{a + b + c} \geq 20$, but I didn't know how to continue from here.

Do you have any suggestions for this inequality?

Answer 1

Your first step gives a wrong inequality. Try $c\rightarrow0+$ and $a=b=1$.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Thus, we need to prove that $$\frac{\sum\limits_{cyc}a^2(a+b)(a+c)}{9uv^2-w^3}+\frac{12}{u}\geq20.$$ Now, we see that it's a linear inequality of $w^3$ because $\sum\limits_{cyc}a^2(a+b)(a+c)$ is fourth degree and the condition does not depend on $w^3$.

Indeed, $$\sum_{cyc}a^2(a+b)(a+c)=\sum_{cyc}a^2(a(a+b+c)+bc)=$$ $$=(a^3+b^3+c^3)(a+b+c)+(a+b+c)abc=(27u^3-27uv^2+3w^3)3u+3uw^3.$$ Hence, we need to prove that $$\frac{(27u^3-27uv^2+3w^3)3u+3uw^3}{9uv^2-w^3}+\frac{12}{u}\geq20,$$ which is a linear inequality of $w^3$ after full expanding.

Thus, it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.

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

Let $c\rightarrow0^+$.

Thus, we need to prove that $$\frac{a^2}{\frac{1}{a}}+\frac{\frac{1}{a^2}}{a}+\frac{36}{a+\frac{1}{a}}\geq20$$ or $$(a-1)^4(a^4+4a^3+11a^2+4a+1)\geq0;$$ 2. Two variables are equal.

Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$.

Thus, we need to prove that: $$\frac{2a^2}{a+\frac{1-a^2}{2a}}+\frac{\left(\frac{1-a^2}{2a}\right)^2}{2a}+\frac{36}{2a+\frac{1-a^2}{2a}}\geq20$$ or $$(1-a)(1+a+3a^2-157a^3+415a^4-225a^5+381a^6-99a^7)\geq0,$$ which is smooth.

We can use also the following way. $$\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2(ab+ac+bc)}{b+c}=a^3+b^3+c^3+\sum_{cyc}\frac{a^2bc}{b+c}=$$ $$=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc+\sum_{cyc}\frac{a^2bc}{b+c}\geq$$ $$\geq(a+b+c)^3-3(a+b+c).$$ Id est, it's enough to prove that $$(a+b+c)^3-3(a+b+c)+\frac{36}{a+b+c}\geq20.$$ Can you end it now?

Answer 2

I can easy get an SOS (Sum of Squares)! We have: $$\text{LHS-RHS} =\frac{A}{(a+b)(b+c)(c+a)(a+b+c)}\geqq 0$$ Where: $$A=\Big[4\, \left( a+b+c+5+\sqrt {3} \right) \left( a+b+c-\sqrt {3} \right) +20\,\sqrt {3}-24\Big]abc$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b+c \right) \Big[ \left( a+b+c \right) ^{2 }+4\,a+4\,b+4\,c+9\Big] \left( a+b+c-2 \right) ^{2} \geqq 0$$ Equality holds when $a=b,\,c=0$ and its permutations.

Answer 3

Let $p=a+b+c,\,q=ab+bc+ca,$ we write your inequality as $$ \sum \frac{a^2}{b+c} + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca},$$ or $$(a+b+c) \sum \frac{a}{b+c} - (a+b+c) + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca}.$$ Use $$ \sum \frac{a}{b+c} \ge \frac{a^2+b^2+c^2}{ab+bc+ca} = \frac{p^2}{q}-2.$$ We need to prove $$p\left(\frac{p^2}{q}-2\right) - p + \frac{36q} {p}\geqslant 20\sqrt{q},$$ equivalent to $$\frac{(p^4-3p^2q+36q^2)^2}{p^2q^2} \geqslant 400q,$$ or $$\frac{(p^4+2p^2q+81q^2)(p^2-4q)^2}{p^2q^2} \geqslant 0.$$ Equality occur when $a=b=1,\,c=0.$ The proof is completed.