Need Help

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+zx2xyz727xy + yz +zx -2xyz \le \frac{7}{27}

My approach :

Using the identities:

1.x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx) 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=x2+y2+z2+2xy+2yz+2zx 2. (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx

The given inequality can be rewritten as :

2(x3+y3+z3)32(x2+y2+z2)+1229 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)=2t332t2+16 f(t) = 2t^3 - \frac{3}{2}t^2 + \frac{1}{6}

The above function becomes convex for t 14\ge \frac{1}{4}

So, using Jensen's inequality on a,ba,b and cc

f(a)+f(b)+f(c)3f(13) \frac{f(a)+f(b)+f(c)}{3} \ge f(\frac{1}{3})

We get,

2(x3+y3+z3)32(x2+y2+z2)+122/9 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,yx,y and zz which are greater than 14\frac{1}{4} . I do not have the proof that an yet lower value cannot be achieved when i consider all the x,yx,y and zz which range from 0 to 1.

Note by A Former Brilliant Member
3 years, 9 months ago

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1 vote

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

A Former Brilliant Member - 3 years, 9 months ago

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Try homogenizing the inequality.

Daniel Liu - 3 years, 9 months ago

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