<s>[INST] How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
 [/INST] Set $a^2+b^2=3-c^2=k^2$, then we want to maximize
$$a+b=b+\sqrt{k^2-b^2}$$
Since it is increasing in $k$,  and $k^2=3-c^2\le 3$, we achieve maximum at $k=\sqrt{3}$, $c=0$ (!)
Set $b=\sqrt{3}\sin\alpha \le 1$. Therefore
$$0\le \sin(\alpha)\le\frac{1}{\sqrt{3}}<\frac{1}{\sqrt{2}}\implies 0^{\circ}\le\alpha<45^{\circ}$$
And we want to maximize
$$a+b=\sqrt{3}(\sin\alpha+\cos\alpha)=\sqrt{6}\sin(\alpha+45^{\circ})$$
Since both $b$ and the function are incresing in $\alpha\in[0,45^{\circ}]$ and decreasing in $c$, we reach maximum at $c=0$, $b=1$, $a=\sqrt{2}$.
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