How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$

Question

let $a,b$ are positive numbers,and such $ab=8$ find this minum

$$\sqrt{a^2+64}+\sqrt{b^2+1}$$

My try:

and I find when $a=4,b=2$,then $$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$

it maybe use Cauchy-Schwarz inequality

Thank you

Did you notice that $\sqrt{a^2+64}+\sqrt{b^2+1}=\sqrt{a^2+(ab)^2}+\sqrt{b^2+1}=a\sqrt{1+b^2}+\sqrt{1+b^2}=(a+1)\sqrt{b^2+1}$ ? — Mher, Sep 12 '13 at 06:57

score 6 · Accepted Answer · answered Sep 12 '13 at 08:01

6

by the Cauchy-Schwarz inequality,we have $$(a^2+64)(1+4)\ge(a+16)^2$$ $$(b^2+1)(4+1)\ge (2b+1)^2$$ then $$\sqrt{a^2+64}+\sqrt{b^2+1}\ge\dfrac{1}{\sqrt{5}}(a+2b+17)$$ and By AM-GM,we have $$a+2b\ge 2\sqrt{2ab}=8$$ so $$\sqrt{a^2+64}+\sqrt{b^2+1}\ge 5\sqrt{5}$$

answered Sep 12 '13 at 08:01

math110

93,304

oh,It's very beautiful,Thank you !+1 – math110 Sep 12 '13 at 08:06
This is the best solution so far. – Ewan Delanoy Sep 12 '13 at 08:06

score 3 · Answer 2 · answered Sep 12 '13 at 07:15

3

As noted by Mher Safaryan, your inequality is equivalent to

$$ (1+a)\sqrt{b^2+1} \geq 5\sqrt{5} \tag{1} $$

or in other words,

$$ (1+a)^2(1+b^2) \geq 125 \tag{2} $$

or equivalently,

$$ (1+\frac{8}{b})^2(1+b^2) \geq 125 \tag{3} $$ We are now done, because of the identity

$$ (1+\frac{8}{b})^2(1+b^2)-125=\frac{(b-2)^2\bigg(b^2+20b+16\bigg)}{b^2} $$

answered Sep 12 '13 at 07:15

Ewan Delanoy

61,600

How do you get to the first inequality? The title of the question is a bit confusing, but I don't think there is an inequality given. – Daniel R Sep 12 '13 at 07:32
@DanielR True, the constant $5\sqrt{5}$ is not in the title, but it's in the text. – Ewan Delanoy Sep 12 '13 at 07:37
But the question is to minimize the expression in the title right? $5\sqrt5$ comes from the asker's attempt to solve it, no? – Daniel R Sep 12 '13 at 07:40
Yes, this answer takes advantage of the work already made by the asker. – Ewan Delanoy Sep 12 '13 at 07:45

score 3 · Answer 3 · answered Sep 12 '13 at 07:16

Setting $x = a$ and $b = \frac{8}{x}$ we find that we want to minimise $$\sqrt{x^2 + 64} + \sqrt{\frac{64}{x^2} + 1} = \left (\frac{1}{x} + 1 \right ) \sqrt{x^2 + 64}$$ Differentiating, we obtain $$ \frac{x^3 - 64}{x^2 (x^2 + 64)} $$ Setting this equal to zero, we obtain a real extremum at $x=4$, corresponding to $a=4, b=2$ and a value of $5\sqrt{5}$. Analysis of the second derivative shows that this is a minimum.

How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$

3 Answers3