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.

By $1$-d version of polarization identity $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$,


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*If $a^2 + b^2 = 2$, then
$$(a+b)^2 = 2(a^2+b^2) - (a-b)^2\le 2(a^2+b^2) = 4 \quad\implies\quad a+b \le |a+b| \le 2$$

*If $a + b = 2$, then
$$
8(a^4+b^4) = 4((a^2+b^2)^2 + (a^2-b^2)^2)\\
\ge 4(a^2+b^2)^2 = (2(a^2+b^2))^2 = ((a+b)^2 + (a-b)^2)^2\\
\ge (a+b)^4 = 16\\
{\large\Downarrow}\\
a^4+b^4 \ge 2
$$
