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

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

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TopNewestYes, 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

\[ (a+b-c)(b+c-a)(c+a-b)≤abc \] – Dinesh Chavan · 2 years, 8 months ago

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– Dinesh Chavan · 2 years, 8 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 inequalityLog in to reply

– Mathh Mathh · 2 years, 8 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.Log in to reply

@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 \[a+b-c+b+c-a=2a\] \[b+c-a+c+a-b=2c\] \[c+a-b+a+b-c=2b\]

Nice questionBut none of the above results is negative, since \(a,b,c\) are positive real numbers. Hence, we can say that \((a+b-c)(b+c-a)(c+a-b)≤abc\) – Dinesh Chavan · 2 years, 8 months ago

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