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$
 A: Let $b^2=1-\delta$ where $\delta\ge 0$. Then, $a^2=2-c^2+\delta$. Now, $(a+b)^2=a^2+b^2+2ab=3-c^2+2\sqrt{1-\delta}\sqrt{2-c^2+\delta}$. The product of square roots is decreasing in $\delta$ (we can explicitly differentiate or just note that $1-\delta<1$ while $2-c^2+\delta>1$ so a small increase in $\delta$ decreases first square root by a greater percentage than it increases the second square root). Substituting $\delta=0$, $(a+b)^2\le 3-c^2+2\sqrt{2-c^2}<3+2\sqrt{2}=(1+\sqrt{2})^2$ so $a+b<1+\sqrt{2}$. Note strict inequality if $c>0$.
A: As Sun stated, given a valid tuple of $(a,b,c)$, replace it with $(A, b, 0)$ where $ A^2 = a^2 + c^2$. Observe that $a + b \leq A + b$, hence it remains to show that $ A + b \leq 1 + \sqrt{2}$.
Squaring this, we need to show that $ A^2 + 2Ab + b^2 \leq 3 + 2 \sqrt{2}$ or that $AB \leq \sqrt{2}$.
But since $A^2 + 2b^2 \leq 4$, hence $ 2 \sqrt{2} Ab \leq A^2 + 2b^2 \leq 4$, hence we do have $Ab \leq \sqrt{2}$.
A: 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 $c=0$, $b=1$, $a=\sqrt{2}$.
