# Trigonometry or inequality problem

Today, I saw this question:

If $x,y,z \in [0,\frac\pi 2]$, $x+y+z=\frac{3\pi}{4}$ and $\sec^2(x)\sec^2(y)\sec^2(z)=8$, calculate $E=\tan x\tan y+\tan y\tan z+\tan z\tan x$

My first thought was getting everything in term of $\tan$, and everything went as expected. Let $a=\tan x, b=\tan y, c= \tan z$. Then, after playing with the equations I got:

$$x+y+z=\frac{3\pi}{4}$$ $$\implies 1+a+b+c=ab+bc+ac+abc \tag1$$ And since $\tan^2x+1=\sec^2x$, by AM-GM we have $$8=(1+a^2)(1+b^2)(1+c^2)\ge 8abc$$ $$\implies 1 \ge abc$$ So now I want to show that $$(1+a^2)(1+b^2)(1+c^2)=8 \implies a+b+c\ge ab+bc+ac$$ in order to show that $LHS\ge RHS$ , so that they only achieve equality when $a=b=c=1$. Wolfram confirms that it is possible to do so, but I have not managed to prove it.

Edit: A solution to the problem

From $(1)$, we know: $$a=\frac{(b+c)+(1-bc)}{(b+c)-(1-bc)}$$ Substituting $u=b+c$, $v=1-bc$, we have that $u^2+v^2=(1+b^2)(1+c^2)$ (A special case of the Brahmagupta-Fibonacci identity). Therefore $$1+a^2=2\frac{u^2+v^2}{(u-v)^2}=2\frac{(1+b^2)(1+c^2)}{(u-v)^2}=\frac{8}{(1+b^2)(1+c^2)}$$ Since $x\le \frac{\pi}{2}$, we have that $y+z\ge \frac{\pi}{4}\implies \tan(y+z)=\frac{u}{v}\ge1$. Therefore, $u-v\ge 0$, and we can safely take the square root without losing solutions.

$$\implies (1+b^2)(1+c^2)=2(u-v)$$ $$\implies (1+b^2)(1+c^2)=2(b+c+bc-1)$$ $$\implies 1+b^2+c^2+b^2c^2=2b+2c+2bc-2$$ $$\implies (b-1)^2+(c-1)^2+(bc-1)^2=0$$ $\implies a=b=c=1, \implies E=3$

But that last inequality I pointed still intrigue me. Can we prove it with elementary methods? Also, is there a simple solution to the problem?

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