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 $(2,4/3)\in K$.
For $a\geqq 1$ $$a^{\,2}+ b^{\,2}\leqq 5$$ $$\because\,3(\,2\,a- 1\,)^{\,2}(\,5- a^{\,2}- b^{\,2}\,)= ab[\,2(\,3\,a- 1\,)(\,b+ 1\,)+ 2- b\,]F+$$ $$+ 4(\,a- 1\,)+ (\,a- 1\,)(\,3- 2\,a\,)[\,(\,a- 1\,)(\,2\,a+ 5\,)+ 2\,]\geqq 0$$