Minimum of $\frac{x^2 + 7}{\sqrt{x^2 + 6}}$ without using calculus?

Question

If $x \in \mathbb{R} $, find the smallest value of $y=\frac{x^2 + 7}{\sqrt{x^2 + 6}}$

Attempted solution:

$y$ can be expressed as $ y=\frac{(x^2 + 6)+1}{\sqrt{x^2 + 6}} = \sqrt{x^2+6} + \frac{1}{\sqrt{x^2 + 6}}$.

However, we can't use AM-GM here since $\sqrt{x^2+6} = \frac{1}{\sqrt{x^2 + 6}}$ when $x^2=-5$.

Using graphing software, I can tell that $y$ minimizes when $x=0,$ I know I can also get the derivative of $y$ with respect to $x$ and get the critical points.

My question is, how can we get the minimum of $y$ analytically (without using calculus)?

Note: Since $y$ is an even function, I think we only need to prove that the function is monotonically increasing on the interval $[0,\infty)$. However, I don't know how to show this without resorting to calculus.

$t+\frac 1 t \geq 2$ for $t>1$ and the value $2$ is attained when $t=1$. — geetha290krm, Jun 26 '23 at 05:04
"don't know how to show this without resorting to calculus" $;-;$ Hint: $;\left(a+\frac{1}{a}\right)$ $- \left(b + \frac{1}{b}\right)$ $= (a-b)(1 - \frac{1}{ab})$ has the same sign as $a-b$ when $a,b \ge 1$. — dxiv, Jun 26 '23 at 05:29
You don't need to prove monotonicity. Let $a=x^2\ge0.$ Then, $6(a+6)\left(y^2-\frac{7^2}6\right)=6(a+7)^2-7^2(a+6)=a(6a+35)\ge0.$ — Anne Bauval, Jun 26 '23 at 06:05
Out of curiosity, how is "analytically" different from "using calculus"? — Anne Bauval, Jun 26 '23 at 06:08

score 4 · Answer 1 · answered Jun 26 '23 at 06:28

You're on the right track: essentially as you do, let $u=\sqrt{x^2+6}$. Then your expression is equal to $${u^2+1\over u}=u+{1\over u}.$$ It's easier to show that this is increasing in $u$ for $u$ in the "appropriate" range, namely $u\in [6,+\infty)$ (which gives us the same result).

To see this, we compute $$u+h+{1\over u+h}-u-{1\over u}=h-{u+h-u\over u(u+h)}=h-{h\over u(u+h)}.$$ Since $u\ge 6$, if $h>0$ we have ${h\over u(u+h)}<h$ as hoped for.

Note that the above argument works if we replace $6$ by any number $>1$.

score 3 · Answer 2 · answered Jun 26 '23 at 06:30

3

If $x=0$ we obtain a value $\frac{7}{\sqrt6}.$

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

Indeed, we need to prove that: $$\frac{x^2+7}{\sqrt{x^2+6}}\geq\frac{7}{\sqrt6}$$ or $$6(x^2+7)^2\geq49(x^2+6)$$ or $$6x^4+35x^2\geq0$$ and we are done.

answered Jun 26 '23 at 06:30

Michael Rozenberg

194,933

1

This was essentially my comment, now edited as an answer. – Anne Bauval Jun 26 '23 at 07:07

score 2 · Accepted Answer · answered Jun 26 '23 at 07:02

2

As commented earlier: you don't need to prove the monotonicity; let $a=x^2≥0.$ Then, $$6(a+6)\left(y^2-\frac{7^2}6\right)=6(a+7)^2-7^2(a+6)=a(6a+35)\ge0$$ hence (since $y\ge0$) the minimum of $y$ is $\frac7{\sqrt6},$ attained for $x=0.$

answered Jun 26 '23 at 07:02

Anne Bauval

34,650

score 0 · Answer 4 · answered Jun 26 '23 at 08:00

From this step we have

$$y=\frac{(x^2 + 6)+1}{\sqrt{x^2 + 6}} = \sqrt{x^2+6} + \frac{1}{\sqrt{x^2 + 6}}>2$$

and by $t= \sqrt{x^2+6}\ge 6$ we have

$$t+\frac1t=m \implies t^2-mt+1=0 \implies t=\frac{m+\sqrt{m^2-4}}{2}$$

which is strictly increasing with $m$, then the minimum is attained for $t=6\iff x=0$.

Minimum of $\frac{x^2 + 7}{\sqrt{x^2 + 6}}$ without using calculus?

4 Answers4