<s>[INST] Given $a^2+b^2=2$ prove $a+b\le2$ 
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*Given $a^2+b^2=2$ prove $a+b\le2$

*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
 [/INST] Using AM-GM inequality:
$\displaystyle\frac{a^2+b^2}{2} \geq \sqrt {a^2b^2}$ 
$\displaystyle{a^2+b^2}\geq  {2ab}$ 
We know :
$\displaystyle\ a^2+b^2=2$ 
Equating the above equations, we get  
$\displaystyle\ 2ab \leq 2$ 
Adding two equations above we get:
$\displaystyle\ a^2+b^2+2ab \leq 4$
Taking Square Roots both the sides:
$\displaystyle\ -2\leq a+b\leq 2$  
Therefore,
$\displaystyle\ a+b\leq 2$

$\displaystyle\frac{a^4+b^4}{2}\geq \sqrt {a^4b^4}$
$\displaystyle{a^4+b^4}\geq 2a^2b^2$
$\displaystyle{a^4+b^4}\geq 2(ab)^2$
We can also write it as:
$\displaystyle{a^4+b^4}\geq 2$
Because 
$\ 2\leq 2(ab)^2$
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