Prove that $\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$

Question

Given $a,b,c>0$ and $a^2\ge b^2+c^2$. Prove that $$\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$$

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This is my try:

I expanded the LHS, and I have to show that $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq 2+3\sqrt2$

I used AM-GM: $\displaystyle \frac{a}{\sqrt2b}+\frac{\sqrt2b}{a}\ge 2, \quad \color{red}{\frac{\left (2-\sqrt2 \right )a}{2b}+\frac{\left ( 1-\sqrt2 \right )b}{a}}$

But it's not good. Then I don't know how.

P/s: I have read this but they are different.

Answer 1

We have to show $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq 2+3\sqrt2\quad\color{red}{(1)}$

$(1)\iff \displaystyle \left ( b+c \right )\left ( \frac{a}{bc}+\frac{1}a \right )+\left (\frac b c +\frac c b \right ) \ge2+3\sqrt2$

We know that $\displaystyle {b+c\ge 2\sqrt{bc},\quad \frac{a}{2bc}+\frac{a}{2bc}+\frac{1}a\ge3\sqrt[3]{\frac{a}{4b^2c^2}}\ge 3\sqrt[3]{\frac{\sqrt{2bc}}{4b^2c^2}}=\frac{3}{\sqrt{2bc}}\\\implies \left ( b+c \right )\left ( \frac{a}{bc}+\frac{1}a \right )\ge 3\sqrt2,\quad \left (\frac b c +\frac c b \right ) \ge 2}$

Remember that $a^2\ge b^2+c^2\ge 2bc$. So we have the Q.E.D

The equality happen $\iff a^2=b^2+c^2$ and $b=c$

Answer 2

hint:

$f(a)=\dfrac{a}{bc}+\dfrac{1}{a}$ has min when $a^2 \ge b^2+c^2$ find this min, then LHS $\ge g(b,c)$