# Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$

Prove that, if $$a,b,c$$ are the sides of a triangle and $$s$$ is the semi-perimeter, then $$a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$$.

What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-

$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$ But I couldn't proceed from this.

Another Idea I had was :- $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$$.
$$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$$

But I did not know how to use this here, and would make the calculations a bit messy, especially of the $$\dfrac{36}{35}$$ part present there.

Can Anyone Help me? Thank You.

• Can someone explain the downvote? Commented Jul 21, 2021 at 13:06

Firstly using $$s= \frac{1}{2} (a+b+c)$$ the inequality is equivalent to $$35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$$. Now note that $$\frac{3abc}{a+b+c} = \frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}$$ and so by the HM-GM-AM inequality it follows that $$\frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}} \leq (abc)^\frac{2}{3} \leq \frac{1}{3}(a^2+b^2+c^2)$$ hence we have $$\frac{72abc}{a+b+c} \leq 8(a^2+b^2+c^2)$$. Now using the AM-QM inequality $$\frac{a+b+c}{3} \leq \sqrt{\frac{a^2 + b^2 + c^2}{3}} \Rightarrow 9(a+b+c)^2 \leq 27(a^2+b^2+c^2)$$. Adding the two inequalities gives $$35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$$ as required.

After expanding, the expression is equivalent to $$13(a^3+b^3+c^3)+4(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)-63abc\geq 0$$

$$\Leftrightarrow 13[a^3+b^3+c^3-3abc]+4[(a^2b+b^2c+c^2a)-3abc]+4[(ab^2+bc^2+ca^2)-3abc]\geq 0,$$ which is clear, as follows from AM-GM.

[EDIT] With more thoughts and less work, it suffices by AM-GM to prove a stronger result: $$a^2+b^2+c^2\geq \frac{36}{35}\left(s^2+\frac{(2s/3)^3}s\right),\qquad (1)$$ where one replaces $$abc$$ by $$((a+b+c)/3)^3=(2s/3)^3\geq abc.$$

Clearly (1) is equivalent to $$3(a^2+b^2+c^2)\geq (a+b+c)^2,$$ which is true by C-S.

• Nice Solution! (assuming you expanded it correctly). Can you tell me how did you know you will get some sort of a way out when you expanded everything, like some sort of a motivation? As far as I know expanding in Inequalities always get messy, and usually do not lead to the solution. Commented Jul 21, 2021 at 12:36
• @Anonymous Thanks. You are right in this: "As far as I know expanding in Inequalities they always get messy". Exploiting the symmetry is what I usually try first. I can only say that we get lucky in this problem, because it all depends on AM-GM, and no other tricks. Commented Jul 21, 2021 at 12:40
• @Pythagoras About your last post: My idea would be to use the fact that, if $f:]a,b[\to\mathbb R$ is convex, then there exists a $c\in[a,b]$ such that $f$ is monotone decreasing on $]a,c[$ (possibly the empty set) and monotone increasing on $]c,b[$. (Cf. math.stackexchange.com/a/2160953/631742) Commented Jul 24, 2021 at 9:49

Here is another approach.

By AM-GM,

$$(a+b+c)(a^2+b^2+c^2) \geq 3 (abc)^{2/3} \cdot 3 (abc)^{1/3} = 9abc$$
$$4 (a^2+b^2+c^2) \geq \cfrac{18abc}{s}$$
$$13 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + \cfrac{18abc}{s}$$

Using $$a^2+b^2+c^2 \geq ab+bc+ca$$,

$$13 (a^2+b^2+c^2) \geq 9(ab+bc+ca) + \cfrac{18abc}{s}$$
$$26 (a^2+b^2+c^2) \geq 18(ab+bc+ca) + \cfrac{36abc}{s}$$
$$35 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + 18(ab+bc+ca) + \cfrac{36abc}{s}$$
$$35 (a^2 + b^2 + c^2) \geq 36 s^2 + \cfrac{36abc}{s}$$

$$a^2 + b^2 + c^2 \geq \cfrac{36}{35} \left(s^2 + \cfrac{abc}{s} \right)$$

It's true for any positives $$a$$, $$b$$ and $$c$$.

Indeed, we need to prove that: $$a^2+b^2+c^2\geq\frac{36}{35}\left(\frac{(a+b+c)^2}{4}+\frac{2abc}{a+b+c}\right)$$ or $$35(a^2+b^2+c^2)(a+b+c)\geq9(a+b+c)^3+72abc$$ or $$\sum_{cyc}(13a^3+4a^2b+4a^2c-21abc)\geq0,$$ which is true by Muirhead or by AM-GM.

The best result in this type it's the folowing.

Let $$a$$, $$b$$ and $$c$$ be positive numbers. Prove that: $$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)}\geq\frac{(a+b+c)^2}{2}.$$