TITLE: How to prove this inequality $a+b+c\ge \sqrt{3}+\frac{1}{4}c^2(a-b)^2$
QUESTION [8 upvotes]: 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

REPLY [6 votes]: 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.