<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] 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 $c>0$.
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