0

If $x > 0, y > 0, z > 0$ and $x+y+z = 1$ then prove $$\dfrac{x}{2x+1}+\dfrac{y}{3y+1}+\dfrac{z}{6z+1}\le \dfrac{1}{2}$$

I saw this post half an hour ago and after coming up with a proof, the post was deleted. So I thought maybe OP of that post wants to see the answer. I decided to post the answer here for those inequality lovers who want to see some cool tricks....I present an answer, but still wanna see others' answers as well. I still learn it from @Michael Rozenberg expertise, but he's not actively posting answers any more. So the inequality community at MSE here have one less active members...

nonuser
  • 90,026

2 Answers2

2

We prove: $\dfrac{3x}{6x+3} + \dfrac{2y}{6y+2} + \dfrac{z}{6z+1} \le \dfrac{1}{2}$.

Put $a = 6x+3, b = 6y+2, c= 6z+1 \implies a+b+c = 12$, and $$3x = \dfrac{a-3}{2}, 2y = \dfrac{b-2}{3}, z = \dfrac{c-1}{6}.$$ So we prove: $$\dfrac{a-3}{2a}+\dfrac{b-2}{3b} +\dfrac{c-1}{6c} \le \dfrac{1}{2}\iff S=\dfrac{3}{2a}+\dfrac{2}{3b}+\dfrac{1}{6c} \ge \dfrac{1}{2}.$$ We have: $$S = \dfrac{3^2}{6a}+\dfrac{2^2}{6b} + \dfrac{1^2}{6c}\ge \dfrac{(3+2+1)^2}{6a+6b+6c}= \dfrac{36}{6\cdot 12}=\dfrac{1}{2}.$$ Done !

Patricio
  • 1,604
1

Lemma: For all $x>0$ and $a>0$ we have $${x\over ax+1}\leq {1\over 4}x +{1\over 4a}$$

Proof: Left to the reader.

So $$ {x\over 2x+1} + {y\over 3y+1} +{z\over 6z+1} \leq {1\over 4}((x+y+z) +{1\over 2}+{1\over 3} +{1\over 6} ) = {1\over 2} $$

nonuser
  • 90,026