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

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

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}

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

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.

• -1 This is incorrect. What you have obtained is a lower bound and not the minimum. The minimum value is $2+\sqrt2$. Apr 16 '14 at 18:16
• @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 $\sqrt{2}$. Apr 16 '14 at 18:49
• @Jibarito The Blumenthal's solution is total wrong. See my solution. Apr 8 '17 at 17:06

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!

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 $a + b \geq \sqrt{2}$.

• 2ab < 1. Your last line is wrong. Apr 16 '14 at 18:54
• @LAcarguy I don't understand- you can take $a = b = 1/\sqrt{2}$, in which case $2 a b = 1$ holds. Apr 16 '14 at 19:03
• @A Blumenthal $a+b\geq\sqrt2$ is wrong. Try $a\rightarrow1$ and $b\rightarrow0^+$. Apr 8 '17 at 16:34
• @MichaelRozenberg I completely agree... not sure what I was thinking. Apr 8 '17 at 20:13
• @A Blumenthal See please my proof. Apr 8 '17 at 20:16

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 $x = \frac{\pi}4$.

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