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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$

  • Show that c = 0 improves a+b; Then evaluate $\max 3 +a - a^2$ in the interval $a \in [\sqrt(2) \sqrt(3)]$; This function is strictly decreasing in that interval, so $a = \sqrt(2)$ will give the largest $a+b = 1 + \sqrt(2)$. – sredni vashtar Oct 08 '13 at 14:35
  • @Sun If we let $c$ to be 0, we can easily achieve equality. But apparently, that is not allowed. – chubakueno Oct 08 '13 at 14:37
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    @chubakueno I am trying to say that making c > 0 can only make the remaining terms smaller - we can formally prove this no? – sredni vashtar Oct 08 '13 at 14:40

3 Answers3

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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 $c>0$.

user96614
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1

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 $Ab \leq \sqrt{2}$.

Calvin Lin
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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 $c=0$, $b=1$, $a=\sqrt{2}$.

chubakueno
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