# My Good Friend Ravi

Let $$a,b,$$ and $$c$$ be positive reals. Prove $abc\geq (a+b-c)(b+c-a)(c+a-b)$.

Note by Joshua Ong
3 years, 11 months ago

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Yes, one of my favourite substitution. But please note that for getting help from your best friend Ravi Substitution, shouldn't $$a,b,c$$ be given as sides of a triangle. Anyways, I will post a proof

Lets assume $$a,b,c$$ are sides of a triangle. Then By using Ravi Substitution,$a=x+y,b=y+z,c=z+x$ So, now it is left to prove that $(x+y)(y+z)(z+x) \geq 8xyz$ Now, its very easy using AM-GM, As $x+y \geq 2 \sqrt{xy},y+z \geq 2 \sqrt{yz},z+x \geq 2 \sqrt{zx}$ Multiplying the above results, we get out desired result

Now, Let's assume that $$a,b,c$$ are not lengths of a triangle.

Then, at least one of $$(a+b-c),(b+c-a),(c+a-b)$$ terms is negative. So,

$$(a+b-c)(b+c-a)(c+a-b)<0$$

But we have also been given that all $$a,b,c$$ are positive real numbers. Therefore their product $$abc$$ is always positive. Hence
$(a+b-c)(b+c-a)(c+a-b)≤abc$

- 3 years, 11 months ago

Also, you can get another solution by expanding. It becomes $\displaystyle \sum_{\text{cyc}} a^3 + 3abc \ge \displaystyle \sum_{\text{sym}} a^2b$ Which is now a direct application of Schur's inequality

- 3 years, 11 months ago

But wait, you're saying that at least one of $$d,e,f$$ being negative implies $$def<0$$ but of course this is not true when $$2$$ of them are negative.

- 3 years, 11 months ago

Nice question @mathh mathh . We have to show that any two of $$(a+b-c),(b+c-a),(c+a-b)$$ cannot be negative. First lets assume that 2 of them are negative. So, if two of them are negative, then their sum must also be negative. but we see that $a+b-c+b+c-a=2a$ $b+c-a+c+a-b=2c$ $c+a-b+a+b-c=2b$

But none of the above results is negative, since $$a,b,c$$ are positive real numbers. Hence, we can say that $$(a+b-c)(b+c-a)(c+a-b)≤abc$$

- 3 years, 11 months ago