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?

  • $\begingroup$ Where is this problem from? $\endgroup$ – TBTD Jan 4 at 15:54
  • $\begingroup$ The problem is from the "Mathematics Magazine": gmb.ssmr.ro $\endgroup$ – Sandel Jan 4 at 16:07

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?

  • $\begingroup$ In the first inequality, I get $9uv^2-\mathbf{3}w^3$ in the denominator of the first term, and ${\mathbf{12}\over u}$ for the second term. Also, would you please add some details on how you deduce that this is a linear inequality in $w^3?$ I don't follow the explanation at all. $\endgroup$ – saulspatz Jan 4 at 16:53
  • $\begingroup$ @saulspatz In first it should be $9uv^2-w^3$. For the second it was typo. I fixed. Thank you! The expression $\sum\limits_{cyc}a^2(a+b)(a+c)$ by fourth degree. $\endgroup$ – Michael Rozenberg Jan 4 at 16:55
  • $\begingroup$ Then did you mean to define $w^3=3abc?$ The post says $w^3=abc.$ I know that the numerator is of fourth degree, but I don't see what that has to do with the conclusion that the inequality is linear in $w^3.$ $\endgroup$ – saulspatz Jan 4 at 17:06
  • $\begingroup$ I used the second proof, which is very nice and a bit simpler than the first one. Thank you a lot! $\endgroup$ – Sandel Jan 4 at 17:11
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    $\begingroup$ @MichaelRozenberg : I was almost about to say, where is arqady, then I realized I am not on AoPS, then I saw you already posted ;) $\endgroup$ – TBTD Jan 4 at 18:53

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