0
$\begingroup$

Prove: if $$a+b+c=\frac1a+\frac1b+\frac1c,\quad0

All I tried was $$2a+b+c=a+\frac1a+\frac1b+\frac1c$$ $$LHS=a^2b^2c^2\left(\frac1{((a^2+1)bc+ca+ab)^2}+\frac1{((b^2+1)ca+ab+bc)^2}+\frac1{((c^2+1)ab+bc+ca)^2}\right)$$ that didn't help at all. Thanks.

  • 0
    where did you find this bad boy :) ? – 2017-01-14
  • 0
    @A.Molendijk Somebody asked this question to me. I don't know the source either. – 2017-01-14

1 Answers 1

1

The following solution is an exact copy of the one given on page 396 of the book Olympiad Inequalities, GIL Publishing House. The book also says that this problem was on the shortlist for the International Mathematical Olympiad 2009.

From condition we have $3(ab+ac+bc)=3abc(a+b+c)\leq (ab+ac+bc)^2\Longrightarrow ab+ac+bc\geq 3.$ From Am-Gm inequality it follows: $(2a+b+c)^2\geq 4(a+b)(a+c)$, so it's enough to prove: $$\sum\frac{1}{(a+b)(a+c)}\leq\frac{3}{4} \\ \Longleftrightarrow 8(a+b+c)\leq 3(a+b)(a+c)(b+c)$$ Using condition, we can see that it's enough to prove: $$8(a+b+c)(ab+ac+bc)\leq 9(a+b)(a+c)(b+c)$$ What is well known inequality and equivalent to: $c(a-b)^2+b(a-c)^2+c(a-b)^2\geq 0$. Equality holds when $a=b=c=1$.