Given $a^2+b^2=2$ prove $a+b\le2$ 
*

*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.
 A: Courtsey : $\rightarrow$ Cauchy Schwarz
$$(a+b)^2\le(1+1)(a^2+b^2)\\\implies a+b\le |a+b|\le 2$$
$$$$Now by the fact that $a+b=2$ and $a^2+b^2\ge 2ab$, we deduce that $2(a^2+b^2)\ge 4\implies a^2+b^2\ge 2$ $$(a^2+b^2)^2\leq2(a^4+b^4)\\ \implies a^4+b^4\ge 2$$
BINGO!
A: 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 
$\ 2\leq 2(ab)^2$
A: By $1$-d version of polarization identity $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$,


*

*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
$$
8(a^4+b^4) = 4((a^2+b^2)^2 + (a^2-b^2)^2)\\
\ge 4(a^2+b^2)^2 = (2(a^2+b^2))^2 = ((a+b)^2 + (a-b)^2)^2\\
\ge (a+b)^4 = 16\\
{\large\Downarrow}\\
a^4+b^4 \ge 2
$$
A: Let $$a = \sqrt2 \sin(t), \qquad b = \sqrt2 \cos(t)$$
Now we need to show that $$\sin(t) + \cos(t) \le \sqrt2$$ which is equivalent to showing that $$\frac{1}{\sqrt2}\ sin(t) + \frac{1}{\sqrt2}\cos(t) \le 1$$ but $$\frac{1}{\sqrt2}\ sin(t) + \frac{1}{\sqrt2}\cos(t) = sin(\frac{\pi}{4} + t) \le 1$$
