3

Question -

Let $x, y, z$ be distinct real numbers. Prove that $$ \frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1 $$

My work -

first i apply directly C-S and after simplification i have to prove that $4(xy+yz+zx)>x^2+y^2+z^2$ which i am not able to prove ..

then multiplying $x^2,y^2,z^2$ to numerators and denominators of corresponding fractions respectively i again apply C-S and this time we have to prove that after simplification

$(xy)^2+(yz)^2+(zx)^2 + 2x^3y+2y^3z+2z^3x > 0$ which again i fail to prove ...

I try to do with other inequalities but none of them working.

any help will be helpful

thankyou

Ishan
  • 1,725

1 Answers1

6

Let $a = \dfrac{x}{x-y}, b = \dfrac{y}{y-z}, c = \dfrac{z}{z-x}$, and note $a+b+c = ab+bc+ca+1$. We need to show $a^2+b^2+c^2 \geqslant 1$. But this is now just equivalent to $(a+b+c-1)^2 \geqslant 0$

Alternately, the original inequality $\iff \dfrac{(x^2y+y^2z+z^2x-3xyz)^2}{(x-y)^2(y-z)^2(z-x)^2}\geqslant 0$

Macavity
  • 46,381
  • thanks macavity, but how did you notice this fact $a+b+c = ab+bc+ca+1$ ??? i mean how you checked it because direct checking is going very long... – Ishan May 01 '20 at 16:01
  • Sometimes a few computations have to be tried, especially to figure which substitutions to use. It isn't so bad once you account for cyclic symmetry, or you get used to some standard ones. There is an approach with CS inequality also, though arguably that's more troublesome in recognising the relevant terms to simplify. – Macavity May 01 '20 at 16:09
  • and the two things which i mention in question that i have got by C-S ..can they be shown true ???(there is an equal sign also in both) – Ishan May 01 '20 at 16:10
  • The two inequalities you mention don't hold true always. Try $x=-y=1, z=0$ for instance. – Macavity May 01 '20 at 16:12