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Hi , I was trying out one of the IMO 1984 problems :

If \(x,y\) and \(z\) are non-negative real numbers such that \( x+y+z=1\) , prove the following inequality

\(xy + yz +zx -2xyz \le \frac{7}{27} \)

My approach :

Using the identities:

\( 1. x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\)

\( 2. (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx\)

The given inequality can be rewritten as :

\( 2(x^3+y^3+z^3) - \frac{3}{2}(x^2+y^2+z^2) +\frac{1}{2} \ge \frac{2}{9} \)

Consider the following function:

\( f(t) = 2t^3 - \frac{3}{2}t^2 + \frac{1}{6} \)

The above function becomes convex for t \(\ge \frac{1}{4} \)

So, using Jensen's inequality on \(a,b\) and \(c\)

\( \frac{f(a)+f(b)+f(c)}{3} \ge f(\frac{1}{3})\)

We get,

\( 2(x^3+y^3+z^3) - \frac{3}{2}(x^2+y^2+z^2) +\frac{1}{2} \ge 2/9 \)

So, I have proved the inequality . I want to confirm if my approach is correct. I think that it is not the complete proof as this proof included only those \(x,y\) and \(z\) which are greater than \(\frac{1}{4}\) . I do not have the proof that an yet lower value cannot be achieved when i consider all the \(x,y\) and \(z\) which range from 0 to 1.

Note by Mayank Singhal
1 month ago

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@Calvin Lin , @Chew-Seong Cheong, @Daniel Liu ,@Pi Han Goh can u please help me out with this .

Mayank Singhal - 1 month ago

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Try homogenizing the inequality.

Daniel Liu - 1 month ago

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