# If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$ [duplicate]

Let $$a$$, $$b$$ and $$c$$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$

I am very new and novice at this problem. I did a little try but couldn't succeed because I was unable to substitute the left term of the inequality into formula. I know a formula that

$$a^\text{3} + b^\text{3} + c^\text{3} - 3abc$$ = $$(a+b+c)(a^\text{2} + b^\text{2} + c^\text{2} - ab -bc - ca)$$. But how to use this formula in that case isn't known to me. And how to show the relation of the both side and when they will become equal?

A small help will be enough for me. Thanks in advance.

## marked as duplicate by Arnaud D., YuiTo Cheng, Jendrik Stelzner, metamorphy, Yanior WegJun 6 at 17:37

• This is true for any non-negative numbers $a,b,c$ which do not have to side of triangles. Look up this for reference. – dezdichado Feb 18 at 17:21
• In the limiting case $a+b=c$ this becomes $2\leq3.$ I wonder if that's supposed to be a $2$ on the right-hand side. – saulspatz Feb 18 at 17:25
• @dezdichado How did you derive the OPs inequality from Schur's inequality as formulated in the link? – Angela Richardson Feb 18 at 17:26
• If I haven't made a mistake, when a,b,c are the sides of a triangle it can be shown that $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)>2abc$ The two sides are equal when $a+b=c$ and I differentiated the difference w.r.t $c$ assuming $a+b<c.$ – saulspatz Feb 18 at 17:47
• @AngelaRichardson just rearrange the inequality and use Schur for when $n = 1.$ – dezdichado Feb 18 at 19:36

Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that $$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:
$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt{x^6y^6z^6}=6xyz$$
• One only needs two variable AM/GM: $x^2y+yz^2\ge 2xyz$ etc. – Lord Shark the Unknown Feb 18 at 17:49
• I use that $$\frac{a+b+c+d+e+f}{6}\geq \sqrt{abcdef}$$ for $$a,b,c,d,e,f\geq 0$$ – Dr. Sonnhard Graubner Feb 18 at 18:10
It's true for all non-negatives $$a$$, $$b$$ and $$c$$.
Indeed, since our inequality is symmetric, we can assume that $$a\geq b\geq c$$ and we obtain: $$3abc-\sum_{cyc}a^2(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$