Mathematics Asked on January 1, 2022

I encountered the following problem in my textbook-

Let $a, b, c$ be three arbitrary real numbers. Denote $$ x = sqrt{b^2-bc+c^2}, y = sqrt{c^2-ca+a^2}, z = sqrt{a^2-ab+b^2} $$ Prove that $$ xy+yz+zx ge a^2+b^2+c^2 $$

**Textbook’s Solution:**

Rewrite x,y in the following forms $$ x = sqrt{{3c^2over 4}+left(b-{cover 2}right)^2}, y = sqrt{{3c^2over 4}+left(a-{cover 2}right)^2} $$

According to **Cauchy-Schwarz** inequality, we conclude $$ xy ge {3c^2over 4}+{1over 4}left( 2b-cright)left(2a-c right) $$

which implies

$$ sum_{cyc}xy ge {3over 4}sum_{cyc}c^2 +{1over 4}sum_{cyc}left( 2b-cright)left(2a-c right) = sum_{cyc}a^2. $$

**My Approach:**

From the given values of $x$ and $y$–

$$ xy = sqrt{left( c^2-bc+b^2right) left(c^2-ca+a^2 right)} ge c^2+csqrt{ab}+ab $$

So, $$xy+yz+zx ge sum_{cyc}a^2+sum_{cyc}asqrt{bc}+sum_{cyc}bc$$

And it rests to prove that-

$$ sum_{cyc}a^2+sum_{cyc}asqrt{bc}+sum_{cyc}bc ge sum_{cyc}a^2 $$

$$ sum_{cyc}asqrt{bc}+sum_{cyc}bc ge 0 $$

Now my question is that how to prove this? Or are we done?

And also if both approachs are correct, what is the equality case?

The equality case is not given in the textbook so I ask.

Thanks!

Your first step is wrong.

Try $a=b=c$.

Id est, you got a right inequality, but after a wrong step and the proof of your last inequality is not relevant already.

Answered by Michael Rozenberg on January 1, 2022

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