<s>[INST] 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?
 [/INST] Note that each integer has a unique prime factorisation..
$8=2*2*2$
$$(1+a^2)(1+b^2)(1+c^2)=8=2*2*2=(1+1)*(1+1)*(1+1) \ \ \ \  \ \ (*)$$      
Since all terms on the the $R.H.S$ are prime numbers ...
By Euclid's Lemma
$2$ must divide either $(1+a^2)$ or $(1+b^2)$ or $(1+c^2)$ or all of them
Let us assume $2$ divides $(1+a^2)$
Now we see only $a=1$ satisfies this
The other two can be determined using same way or by  solving simultaneous equations (two variables and two equations)..
So $b=c=1$
$$x=y=z=\frac{\pi}{4}$$
$$E=1+1+1=3$$
(You can also do this by comparing numbers in $(*)$)
Hope you find this easier.
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