Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ . 
Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ (equality: $a= 1, b= 2$).

For $a\leqq 1$, the inequality is clearly true!
For $a\geqq 1, F= a^{-1}+ 2b^{-1}- 2\geqq 0$
$$5- a^{2}- b^{2}= \frac{ab\left ( 2(a- 1)(b+ 1)+ b+ 2 \right )F+ (a- 1)( - 4a^{3}+ 15a- 5)}{(2a- 1)^{2}}\geqq 0$$
But $-4a^{3}+ 15a- 5= (2- a)(4a^{2}+ 8a+ 1)- 7\geqq - 7$, we need to prove $-4a^{3}+ 15a- 5\geqq 0$ . 

Note that if $a=2$ and $b=4/3$ then $0\leq a,\,b\leq 2$, $\displaystyle \frac{1}{a}+ \frac{2}{b}= 2$, but 
$$a^{2}+ b^{2}=\frac{52}{9}> 5.$$
P.S. Actually on the compact set 
$$K:=\left\{(a,b)\in [0,2]\times[0,2]:  \frac{1}{a}+ \frac{2}{b}\geq 2\right\}$$
the continuous function $f(a,b)=a^2+b^2$ attains its maximum value just at $(2,4/3)\in K$.
