abc= prove: $3+2(ab+bc+ca)\ge\sqrt{1+4(a+b)}+\sqrt{1+4(b+c)}+\sqrt{1+4(c+a)}$

Question

Let $a,b,c>0: abc=1.$ Prove that: $$3+2(ab+bc+ca)\ge\sqrt{1+4(a+b)}+\sqrt{1+4(b+c)}+\sqrt{1+4(c+a)}$$

My try to prove stronger one: Let $a,b,c>0: abc=1.$ Prove that: $$3+2(ab+bc+ca)\ge\sqrt{3(3+8(a+b+c))}$$ which is false. Also, I denoted $(a,b,c)=(\frac{x}{y},\frac{y}{z},\frac{z}{x})$ but the rest seems complicated.

I hope someone can help me a hint to solve this hard problem. Thanks!

Remark. It've been here for a long time. Is the following stronger inequality true? $$3+2(ab+bc+ca)\ge\sqrt{3}\cdot\sqrt{3+8(a+b+c)}.$$

Interestingly, the inequality still holds with $a^2+b^2+c^2$ instead of $3$ (Wolfram Alpha). Unfortunately, you need to prove the inequality for a smaller LHS so this doesn't work. — Toby Mak, Dec 15 '21 at 09:42
@TobyMak $ (a+b+c)^2 \geq 3 + 2(ab+bc+ca) \geq \sum \sqrt{ 1+ 4(a+b) }$. Your inequality is weaker than OP's. — Calvin Lin, Dec 15 '21 at 15:44
@Mars, OP means “original poster” or “original post”. It means you or this question here. — Apass.Jack, Dec 16 '21 at 10:43
@Apass.Jack I got it. Is there any hint for me? It seems nightmare to prove it. The problem is nice but I don't see any idea to solve it. — Sickness, Dec 16 '21 at 12:14
I should have said that $3+2(ab+bc+ca)\ge\sqrt{3(3+8(a+b+c))}$ is true. — Toby Mak, Dec 17 '21 at 13:24
@TobyMak, did you say "$3+2(ab+bc+ca)\ge\sqrt{3(3+8(a+b+c))}$ is true"? I meant that inequality is not true. — Apass.Jack, Dec 17 '21 at 13:57

score 0 · Answer 1 · answered Dec 18 '23 at 15:38

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Proof.

By using AM-GM \begin{align*} \sqrt{1+4(b+c)}&= 2\cdot\sqrt{\frac{1}{4}+abc(b+c)}\\&\le \frac{\dfrac{1}{4}+abc(b+c)}{bc+\dfrac{1}{2}}+bc+\dfrac{1}{2}\\&=\frac{\dfrac{1}{4}+abc(b+c)}{bc+\dfrac{1}{2}}+bc-\dfrac{1}{2}+1\\&=\frac{bc(bc+ca+ab)}{bc+\dfrac{1}{2}}+1\\&=2(ab+bc+ca)\cdot\frac{1}{a+2}+1. \end{align*} Thus, it is sufficient to prove $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le 1$$ $$\iff \frac{ab+bc+ca+4(a+b+c)+12}{abc+2(ab+bc+ca)+4(a+b+c)+8}\le 1$$ $$\iff ab+bc+ca\ge 3, $$which is obvious.

Hence, the proof is done. Equality holds iff $a=b=c=1.$

answered Dec 18 '23 at 15:38

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Very elegant proof! Thank you so much. – Sickness Dec 19 '23 at 04:00
@Sickness why do you not accept it ? – mick Dec 23 '23 at 21:37

abc= prove: $3+2(ab+bc+ca)\ge\sqrt{1+4(a+b)}+\sqrt{1+4(b+c)}+\sqrt{1+4(c+a)}$

1 Answers1