# Prove $\min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$

Prove that $$\min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$ Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$ Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4$.

I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?

• Lagrange multipliers quickly tells you that $a=b$ at the min.
– lulu
Commented Jul 13, 2018 at 19:28
• This is a classic problem in Lagrange optimization, i.e., optimizing $f(a,b) = a + b + 1/a + 1/b$ subject to the constraint $g(a,b) = a^2 + b^2 - 1 = 0$. As @lulu points out, symmetry immediately shows that $a=b$ and the solution just drops out. Commented Jul 13, 2018 at 19:29
• thx all for the hints, I'm not that familiar with Lagrange staff but I'll expand my knowledge on it. Commented Jul 13, 2018 at 19:33
• @DavidG.Stork Why can't there be another stationary point than the symmetric solution Commented Jul 13, 2018 at 19:34
• @DavidG.Stork, I see sometimes people use the symmetry argument to prove the max value and here the min value, what is the criterion? Commented Jul 13, 2018 at 19:43

Notice \begin{align} a + b + \frac1a + \frac1b = &\; \left(a + \frac{1}{2a} + \frac{1}{2a}\right) + \left(b + \frac{1}{2b} + \frac{1}{2b}\right)\\ \\ \color{blue}{\rm AM \ge \rm GM \rightarrow\quad} \ge &\; 3\left[\left(\frac{1}{4a}\right)^{1/3} + \left(\frac{1}{4b}\right)^{1/3}\right]\\ \color{blue}{\rm AM \ge \rm GM \rightarrow\quad} \ge &\; 6 \left(\frac{1}{16ab}\right)^{1/6}\\ \color{blue}{a^2 + b^2 \ge 2 ab \rightarrow\quad} \ge &\; 6 \left(\frac{1}{8(a^2+b^2)}\right)^{1/6}\\ = &\; 6 \left(\frac18\right)^{1/6} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \end{align} Since the value $3\sqrt{2}$ is achieved at $a = b = \frac{1}{\sqrt{2}}$, we have

$$\min \left\{ a + b + \frac1a + \frac1b : a, b > 0, a^2+b^2 = 1 \right\} = 3\sqrt{2}$$

Notes

About the question how do I come up with this. I basically know the minimum is achieved at $a = b = \frac{1}{\sqrt{2}}$. Since the bound $3\sqrt{2}$ on RHS of is optimal, if we ever want to prove the inequality, we need to use something that is tight when $a = b$. If we want to use AM $\ge$ GM, we need to arrange the pieces so that all terms are equal. That's why I split $\frac1a$ into $\frac{1}{2a} + \frac{1}{2a}$ and $\frac1b$ into $\frac{1}{2b} + \frac{1}{2b}$ and see what happens. It just turns out that works.

• thx for the simple solution Commented Jul 13, 2018 at 20:54
• These sort of solutions always amaze me, it looks like magic; can you explain how you came to solve it like this? Or is it just experience?
– Ovi
Commented Jul 13, 2018 at 20:56
• You can apply the AM-GM on the first line directly Commented Jul 13, 2018 at 20:57
• @Wolfdale I know that, I just present the answer in the order I come up with the solution. Commented Jul 13, 2018 at 21:12
• @Ovi it is part of experience, part of you know what the answer should be and by trial an error, find something tight enough to get you what you want. Commented Jul 13, 2018 at 21:14

My answer is a little roundabout but without calculus and without pictures or symmetry:

Arithmetic-geometric inequality: $$a+b \geq 2\sqrt{ab}$$ Harmonic-geometric inequality and some rearrangement: $$\sqrt{ab} \geq \frac{2}{\frac{1}{a}+\frac{1}{b}}$$ $$\frac{1}{a} + \frac{1}{b}\geq \frac{2}{\sqrt{ab}}$$ Add both results to get $$a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)~~~~~~~~~~(1)$$ Also note that by the given constraint $$0\leq(a-b)^2 = a^2+b^2-2ab = 1 - 2ab$$ and therefore $$\sqrt{ab} \leq \frac{1}{2}\sqrt{2}$$ Now let $x:=\sqrt{ab}$.

So $0\leq x \leq \frac{1}{2}\sqrt{2} < 1$

To finish off, all we need to show is that the right-hand side $2(x+\frac{1}{x})$ of inequality $(1)$ is minimal if $x$ is maximal and therefore $\frac{1}{2}\sqrt{2}$ because then $$a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\frac{1}{2}\sqrt{2}+\sqrt{2}\right) = 3 \sqrt{2}$$ is always true.

To show that, let's show that the function $f:x\mapsto x+\frac{1}{x}$ is decreasing on the interval $(0,1)$. That's easy with calculus. Without:

Let's choose $h,x$ arbitrarily such that $0<h<1$ and $0<x<x+h<1$.

Then rearrange equivalently or backwards-implicatively to get to our monotonicity claim from a true statement

$$x+h+\frac{1}{x+h} < x+\frac{1}{x}$$ $$h+\frac{1}{x+h} < \frac{1}{x}$$ Multiply through $$hx(x+h) + x < x + h$$ And since $h+x < 1$: $$\Leftarrow hx + x \leq x +h$$ Since also $x<1$: $$\Leftarrow h + x \leq h+x$$ which is true. Since $x,h$ were arbitrary from $(0,1)$, this proves monotonicity and hence the claim.

We will square the whole inequality $$\frac{(a+b)^2}{(ab)^2}+(a+b)^2+2\frac{(a+b)^2}{ab}\geq 18$$ simplifying and using that $$a^2+b^2=1$$ we get

$$2(ab)^3-13(ab)^2+4ab+1\geq0$$ this is equivalent to $$(2ab-1)((ab)^2-6ab-1)\geq 0$$

Now we have $$a^2+b^2\geq 2ab$$ this is $$ab\le \frac{1}{2}$$

so both factors $$2ab-1,(ab)^2-6ab-1$$ are non posivite, thus their product is non negative.

Without Calculus:

Multiplying both sides of $a+b+\dfrac 1a + \dfrac 1b \ge 3 \sqrt 2$ by $ab$ we get

$$a^2b+b^2a+a+b \ge 3 \sqrt2 ab$$

which factors as

$$(a+b)(ab+1) \ge 3 \sqrt2 ab$$

and squaring both sides yields

$$(a+b)^2(ab+1)^2 \ge 18(ab)^2$$

But $(a+b)^2 = 1+2ab$, and substituting $x = ab$ get

$$(1+2x)(x+1)^2\ge18x^2$$

Furthermore, we know $x \in\left [0, \dfrac 12 \right ]$ because $a^2+b^2 \ge 2ab$, so $\dfrac 12 \ge ab$.

Thus it remains to show that $f(x) = (1+2x)(x+1)^2 - 18x^2 \ge 0$ on $\left[0, \dfrac 12 \right]$. But expanding $f(x)$ and using the rational root theorem we can factor it as $f(x) = 2(x-1/2)(x^2-6x-1)$. Now $(x- 1/2)$ is non-positive on the required interval, and you can find the roots of $x^2-6x-1$ using the quadratic formula to show that it is negative on that interval.

Done!

This is a summary of the arguments and comments above:

$a^2 + b^2 = 1$ and they symmetry argument that $a = b$ shows that $a = b = {1 \over \sqrt{2}}$, and thus $a + b + 1/a + 1/b = 3 \sqrt{2}$.

Calculus confirms it:

$$f(a) = a + 1/a + \sqrt{1 - a^2} + {1 \over \sqrt{1 - a^2}}$$

so

$${df(a) \over da} = -\frac{a}{\sqrt{1-a^2}}+\frac{a}{\left(1-a^2\right)^{3/2}}-\frac{1}{a^2}+1$$

Set this equal to zero and solve for $a$ to find $a = {1 \over \sqrt{2}}$ and the rest follows.

Here's a plot of $f(a)$:

Indeed, the minimum occurs at $a = {1 \over \sqrt{2}}$ and has value $f(a = 1/\sqrt{2}) = 3 \sqrt{2}$.

• Sorry to continue the comments, but what is this symmetry argument? Why would it give a unique minimum/maximum and how do we know which? Commented Jul 13, 2018 at 20:13
• thx for the answer, I'll wait for another solution not based on symmetry. if there isn't one, I'll consider yours the accepted solution. Commented Jul 13, 2018 at 20:14
• Can you elaborate on your “symmetry argument?” A symmetric target function and a symmetric constraint do not guarantee that the extremum is attained at a point with all variables equal. Commented Jul 14, 2018 at 7:25

We have: $\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=(a+b)^2+2(a+b)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2=1+2ab+4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{1+2ab}{(ab)^2}= 5+2ab+\dfrac{2}{ab}+\dfrac{1}{(ab)^2}+\dfrac{2}{ab}= 5+2ab+\dfrac{4}{ab}+\dfrac{1}{(ab)^2}= 1+\left(4+\dfrac{1}{(ab)^2}\right)+ \left(2ab+\dfrac{4}{ab}\right)\ge 2\sqrt{4\cdot\dfrac{1}{(ab)^2}}+(a+b)^2+\dfrac{4}{\dfrac{(a+b)^2}{4}}=\dfrac{4}{ab}+(a+b)^2+\dfrac{16}{(a+b)^2}\ge \dfrac{4}{\dfrac{a^2+b^2}{2}}+\dfrac{((a+b)^2-8)((a+b)^2-2)}{(a+b)^2}+10= 8+10+\dfrac{(a+b)^2-8)((a+b)^2-2)}{(a+b)^2}\ge 18$ because $0<(a+b)^2\le 2(a^2+b^2) = 2\implies a+b+\dfrac{1}{a}+\dfrac{1}{b}\ge \sqrt{18}=3\sqrt{2}$ , with $=$ occurs at $a = b = \dfrac{1}{\sqrt{2}}$

Calling $a = \cos u, b = \sin u\;\;$ we have

$$\left(\cos u + \frac{1}{\cos u}\right)+\left(\sin u + \frac{1}{\sin u}\right)\ge 2\sqrt{\left(\cos u + \frac{1}{\cos u}\right)\left(\sin u + \frac{1}{\sin u}\right)} = 2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}$$

Now examining

$$f(u) = \frac{(\cos^2 u+1)}{\cos u}\frac{(\sin^2 u + 1)}{\sin u}$$

by symmetry considerations, the feasible minimum is at $u = u_0 = \frac{\pi}{4}\;\;$ (because $\sin u_0 = \cos u_0\;$)

giving the value

$$f(\frac{\pi}{4}) = \frac 92$$

then following we have

$$2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}\ge 2\sqrt{\frac 92} = 3\sqrt2$$

• thx for the answer but there is the restriction of not using calculus. Commented Jul 13, 2018 at 21:00
• @Wolfdale Oh sorry!. I will modify it to accomplish that. Commented Jul 13, 2018 at 21:05

### The following solution may be the most essential solution...

Denote $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}, ~~~x \in (0,1).$$

Since $$f''(x)=\dfrac{3-x}{4x^{5/2}}>0,~~~\forall x \in (0,1),$$ hence $$f(x)$$ is a convex function over $$(0,1)$$. Notice that $$a^2, b^2 \in (0,1).$$Therefore, $$a+b+\frac{1}{a}+\frac{1}{b}=f(a^2)+f(b^2) \geq 2f\left(\frac{a^2+b^2}{2} \right)=2f\left(\frac{1}{2}\right)=3\sqrt{2},$$with the equality holding iff $$a^2=b^2$$, namely, $$a=b=\dfrac{\sqrt{2}}{2}.$$

# Solution 1

By $$H_n \leq G_n \leq A_n \leq Q_n$$,

$$a+b+\frac{1}{a}+\frac{1}{b} \ge a+b+\frac{4}{a+b}=(a+b+\frac{2}{a+b})+\frac{2}{a+b} \geq 2\sqrt{2}+\sqrt{\frac{2}{a^2+b^2}} =3\sqrt{2}.$$

# Solution 2

$$a+b+\frac{1}{a}+\frac{1}{b} \geq 2(\sqrt{ab}+\frac{1}{2\sqrt{ab}})+\frac{1}{\sqrt{ab}}\geq 2\sqrt{2}+ \sqrt{2}=3\sqrt{2}.$$