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If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$

This can be easily done by calculas but is there any way to do do this by algebra

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Another way is to use AM-GM while preserving the point of equality, i.e.:

\begin{align} a+b+\frac1{ab} &= a+b+\frac1{2 \sqrt2 ab} + \left(1-\frac1{2\sqrt2}\right)\frac1{ab} \\ &\geqslant \frac3{\sqrt2}+\left(1-\frac1{2\sqrt2}\right)\frac2{a^2+ b^2} \\ &= \frac3{\sqrt2}+\left(1-\frac1{2\sqrt2}\right)\cdot 2 = 2+\sqrt{2} \end{align}

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$$ a + b + \frac{1}{ab} = a + b + \frac{ a^2 + b^2}{ab} \geq a + b + 2 $$

I have used AM-GM ineq:

$$ \frac{a^2 + b^2}{2} \geq ab $$

Remark: IT is still left to show that $a+b \geq \sqrt{2} $ constrained to $a^2 + b^2 = 1 $. See A Blumenthal's solution.

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    $\begingroup$ -1 This is incorrect. What you have obtained is a lower bound and not the minimum. The minimum value is $2+\sqrt2$. $\endgroup$
    – user141421
    Apr 16 '14 at 18:16
  • $\begingroup$ @Lemur This is a good start. You've reduced the problem to showing that the minimum of $a + b$ over the set $a^2 + b^2 = 1, a, b > 0$ is $\sqrt{2}$. $\endgroup$ Apr 16 '14 at 18:49
  • $\begingroup$ @Jibarito The Blumenthal's solution is total wrong. See my solution. $\endgroup$ Apr 8 '17 at 17:06
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If $a=b=\frac{1}{\sqrt2}$ then $a+b+\frac{1}{ab}=\sqrt2+2$.

We'll prove that it's a minimal value.

Indeed, let $a+b=2u$ and $ab=v^2$.

Hence, the condition gives $4u^2-2v^2=1$,

which says that $v^2=\frac{4u^2-1}{2}$ and $4u^2=1+2v^2\leq1+2u^2$, which gives $u\leq\frac{1}{\sqrt2}$

and we need to prove that $$2u+\frac{2}{4u^2-1}\geq\sqrt2+2$$ or $$(1-\sqrt2u)(2\sqrt2-1+4u+2(1-2u^2))\geq0$$ and we are done!

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Edit: The last line is incorrect.

In light of @Lemur's answer, it will suffice to show that $$ a + b \geq \sqrt{2} $$ where $a,b > 0, a^2 + b^2 = 1$ is enforced.

To see this, note that by AMGM, $ab \leq \frac{a^2 + b^2}{2} = \frac{1}{2}$, and so $$ (a + b)^2 = a^2 + b^2 + 2 ab \geq 1 + 2 ab \geq 1 + 1 = 2 $$ Thus $a + b \geq \sqrt{2}$.

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  • $\begingroup$ 2ab < 1. Your last line is wrong. $\endgroup$
    – DeepSea
    Apr 16 '14 at 18:54
  • $\begingroup$ @LAcarguy I don't understand- you can take $a = b = 1/\sqrt{2}$, in which case $2 a b = 1$ holds. $\endgroup$ Apr 16 '14 at 19:03
  • $\begingroup$ @A Blumenthal $a+b\geq\sqrt2$ is wrong. Try $a\rightarrow1$ and $b\rightarrow0^+$. $\endgroup$ Apr 8 '17 at 16:34
  • $\begingroup$ @MichaelRozenberg I completely agree... not sure what I was thinking. $\endgroup$ Apr 8 '17 at 20:13
  • $\begingroup$ @A Blumenthal See please my proof. $\endgroup$ Apr 8 '17 at 20:16
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Let $a = \sin x, b = \cos x$. Then we need to find the minimum of the function $\cos x + \sin x + \sec(x) \csc(x)$ which is $2+ \sqrt{2}$ at $x = \frac{\pi}4$.

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  • $\begingroup$ Well stat looks a lot like caclulus $\endgroup$ Apr 16 '14 at 18:25

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