Question

This is a more complex OLYMPIAD QUESTION, involving PROOF and AM-GM, if you've never taken a mathematical Olympiad don't bother!

DO NOT POST NON-ANSWERS, I will report you

The question:
Let x,y and z be positive real numbers such that xy+yz+xz=3xyz
Prove that x²y+y²z+z²x>=2(x+y+z)-3. (That is the greater than or equal to sign)
Further, in which case do we have equality?

My working so far:
First step is to note that
$[tex]\frac{1}{x} +\frac{1}{y} +\frac{1}{z} =3$[/tex]

The next step is to apply AM-GM, but I am un-familiar with this. Please help me.

Thank you

Answer 1

Answer:

I have done 2 or 3 maths Olympiads

Step-by-step explanation:

A classic way to prove inequalities is using AM-GM inequality. But my approach is different. Here's my proof :

According to an algebraic identity,

x3+y3+z3−3xyz

=(x+y+z)(x2+y2+z2−xy−yz−zx). ..(1)

As x, y & z are positive numbers,

(x+y+z)>0 ....(2)

Now, let's think about another bracket.

x2+y2+z2−xy−yz−zx

=(1/2)∗(2x2+2y2+2z2−2xy−2yz−2zx)

=(1/2)∗(x2−2xy+y2+y2−2yz+z2+z2−2zx+x2)

=(1/2)∗[(x−y)2+(y−z)2+(z−x)2]

Squares of real numbers are never negative.

(1/2)[(x−y)2+(y−z)2+(z−x)2]≥0.

So, x2+y2+z2−xy−yz−zx≥0 ....(3)

From (1), (2) and (3), we can infer that for positive real numbers x, y and z,

x3+y3+z3−3xyz≥0.

x3+y3+z3≥3xyz. .