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2 \times 3
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a_{i-1}
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\frac{2}{3}
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Comments
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
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=2ab+c−a+c+a−b=2cc+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
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
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
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Also, you can get another solution by expanding. It becomes cyc∑a3+3abc≥sym∑a2b Which is now a direct application of Schur's inequality
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.
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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