How to prove this inequality $a+b+c\ge \sqrt{3}+\frac{1}{4}c^2(a-b)^2$

Question

Let $a,b,c\ge 0,ab+bc+ca=1$, show that

(1):$$a+b+c\ge \sqrt{3}$$ (2): $$a+b+c\ge \sqrt{3}+\dfrac{1}{4}c^2(a-b)^2$$

for $(1)$,I have proof,First see that: $$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$ Hence $$a^2+b^2+c^2\ge ab+bc+ca\Longrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge 3(ab+bc+ac)$$ or$$ (a+b+c)^2\ge 3(ab+bc+ca)=3$$

$$\Longrightarrow a+b+c\ge\sqrt{3}$$ But for $(2)$ I can't prove it

Can you add more steps in your proof of (1)? – Ethan Dec 12 '15 at 03:35 — Ethan, Dec 12 '15 at 03:35
I meant before the $ (a+b+c)^{2} $ – Ethan Dec 12 '15 at 03:58 — Ethan, Dec 12 '15 at 03:58
The $\frac14$ seems to be $2 - \sqrt3$ in disguise. – Théophile Dec 12 '15 at 04:05 — Théophile, Dec 12 '15 at 04:05

J.Gudal · Answer 1 · 2015-12-13T14:53:45.193

Suppose we are given $c$.

Let $x=a+b$ and let $y=a-b$.

Note that $y^{2}=x^{2}+4cx-4$ using the condition $ab+bc+ac=1$

Now our inequality becomes:

$x+c \geq \sqrt3+\frac{c^{2}y^{2}}{4}$ which substituting in $y^{2}=x^{2}+4cx-4$ yields:

$0\geq c^{2}x^{2}+x(4c^{3}-4)-4c^{2}-4c+4(3^{1/2})$

Hence this is the inequality we must prove.

Noting from the first part that,

$x\geq\sqrt3 -c$, and also that:

$(a+b)^{2} \geq (a-b)^2 \Rightarrow x^{2}\geq y^{2} \Rightarrow x^{2}\geq x^{2}+4cx-4 \Rightarrow$ $\frac{1}{c}\geq x$.

And so we have the condition:

$\frac{1}{c}\geq x\geq \sqrt3 -c$

And by showing that this interval $[\sqrt3 -c,\frac{1}{c}]$ is contained within the interval of $x$ such that $0\geq c^{2}x^{2}+x(4c^{3}-4)-4c^{2}-4c+4(3^{1/2})$, our inequality will be proven. The interval of $x$ within which $0\geq c^{2}x^{2}+x(4c^{3}-4)-4c^{2}-4c+4(3^{1/2})$ is true is between its roots ,namely:

$[\frac{2(1-c^{3})-2\sqrt{(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)}}{c^2},\frac{2(1-c^{3})+2\sqrt{(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)}}{c^2}]$

Now denote $f(c)=4c^{2}-c(4\sqrt3+1)+4$. Its turning point is at, $c=\frac{4\sqrt3 +1}{8}$, and $f(\frac{4\sqrt3 +1}{8})=0.07..>0$, hence $f(c)$ is positive.

Now,

$4c^{2}-c(4\sqrt3+1)+4>0 \Rightarrow 4c^{2}-4c\sqrt3>c-4 \Rightarrow 4+4c^{3}-4c^{2}\sqrt3>c^{2}-4c+4 \Rightarrow 4c^{6}-8c^{3}+4+4c^{4}+4c^{3}-4c^{2}\sqrt3>4c^{6}+4c^{4}-8c^{3}+c^{2}-4c+4 \Rightarrow 4[(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)]>(c-2+2c^{3})^{2}>0 \Rightarrow \frac{2(1-c^{3})+2\sqrt{(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)}}{c^2}>\frac{1}{c}$

We are now left to show that:

$\sqrt3 -c\geq \frac{2(1-c^{3})-2\sqrt{(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)}}{c^2}$

Note that:

$c^{2}(3c^{4}+1) \geq 0$ hence,

$c^{6}+3c^{4}-4c^{3}-4c^{2}\sqrt{3}+4 \geq 4c^{6}-8c^{3}+4+4c^{4}+4c^{3}-4c^{2}\sqrt3 \Rightarrow 4[(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)] \geq (c^{3}+c^{2}\sqrt3 -2)^{2} \geq 0 \Rightarrow \sqrt3 -c\geq \frac{2(1-c^{3})-2\sqrt{(c^{3}-1)^{2}-c^{2}(-c^{2}-c+\sqrt3)}}{c^2}$

And we are done.

I'm not sure why I put that in as it doesn't really affect the proof. I'll change that. — J.Gudal, Dec 12 '15 at 08:45
@Macavity You are right but in this case: $x^{2}=(a+b)^{2} \geq (a-b)^{2}=y^{2}$ is true since $a,b \geq 0$ — J.Gudal, Dec 12 '15 at 09:29

How to prove this inequality $a+b+c\ge \sqrt{3}+\frac{1}{4}c^2(a-b)^2$

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