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I am wondering if we can Show that $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$?

(I'm particularly interested in $0<a<b<1$ but I don't think restricting $a$ and $b$ here matters)

I think I've seen an answer using polar coordinates, so perhaps that way could be used, but can we avoid polar coordinates too?

So to state the question precisely:

can we show that $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$, and without using polar coordinates?

basically, I am wondering if there is something like (assuming $a<b$:

$a^2 + b^2 > 2a^2$,

and then somehow showing that $2a^2 > 2ab$

I know that the above way cannot work, because $2a^2$ need not be greater than $2ab$, but perhaps there is some similar approach.

Alternatively, an answer saying why and approach like the one above cannot work (if it cannot work).

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  • $\begingroup$ If $a>b$, then we do have $2a^2>2ab$ for your case of interest (where $a>0$). $\endgroup$ – Minus One-Twelfth May 18 at 20:12
  • $\begingroup$ Well, you can apply the Arithmetic - Geometric Mean inequality to $a^2, b^2$ but I'd say that this was entirely equivalent to the $(a-b)^2≥0$ approach. $\endgroup$ – lulu May 18 at 20:14
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Note first that $|ab|=|a||b|\ge ab$ and let $A=|a|, B=|b|$ so that $A^2=a^2, B^2=b^2$. If $A \neq B$, symmetry allows us to choose (rename) so that $A\gt B\gt 0$. The cases of $B=0$ and $A=B$ are straightforward to check.

Then $$a^2+b^2=A^2+B^2=A^2+AB-AB+B^2=A(A+B)-B(A-B)\gt 2AB-B(A-B)\gt 2AB\ge2ab$$

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  • $\begingroup$ Woah! something like this is pretty much exactly what I was hoping to see. Why did we need to define $A= \vert a \vert$ though? is it because if we didn't use absolute values (i.e. if we added $+ab -ab$ instead) it is not necessarily that $aa+ab > 2ab$? $\endgroup$ – user106860 May 18 at 20:53
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    $\begingroup$ @user106860 Inequalities can get tricky if things are negative - here I can be sure that $B(A-B)\gt 0$ because both factors are positive (if I allow $B=0$ I can put in $\ge$). I saves me having to consider cases. $\endgroup$ – Mark Bennet May 18 at 21:48
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enter image description here

The figure shows squares of areas $a^2$ and $b^2$ (with $b > a$), and two squares of area $c^2$.

The "L"-shaped region inside the $b$ square and outside a $c$ square certainly has larger outer dimensions than the "L" outside the $a$ square and inside the other $c$ square. Moreover, since $c<\frac12(a+b)$, the first "L" is also thicker than the second. Consequently, the first "L" has more area than the second, and we conclude $$b^2 - c^2 > c^2 - a^2 \qquad\to\qquad a^2 + b^2 > 2 c^2$$ Since $c = \sqrt{ab}$ (by the classical construction of the geometric mean), the result follows. $\square$

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    $\begingroup$ This is cool. Thank you. $\endgroup$ – user106860 May 18 at 20:57
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You can look at the function

$$f\left(a,b\right)=a^{2}+b^{2}-2ab$$

It has an extermal point when $\nabla f=\left(2a-b,2b-a\right)=\left(0,0\right)$ or $\left(a,b\right)=\left(0,0\right)$. The Hessian is

$$H=\left(\begin{matrix}2&-1\\-1&2\end{matrix}\right)$$

with eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=3$, such that it is positive-definite. Thus $\left(a,b\right)=\left(0,0\right)$ is a minimum and

$$f\left(a,b\right)\geq f\left(0,0\right)$$

or

$$a^{2}+b^{2}-2ab\geq 0$$

as needed.

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  • $\begingroup$ Dude don't want to look at $(a-b)^2 = a^2 + b^2 -2ab$ but wants to look at a Hessian and eigenvalues? Reminded me of this. $\endgroup$ – Billy Rubina May 19 at 3:26
  • $\begingroup$ @BillyRubina Personally, I would stick to $\left(a-b\right)^{2}\geq 0$. $\endgroup$ – eranreches May 19 at 10:10
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Let $$ f(x) = x^2 + b^2 -2bx. $$ Then $$ f'(x) = 2x - 2b = 2(x-b). $$ When $x > b$ this is positive so $f$ is increasing there. Since $f(b)=0$, $f(a) > 0$ when $a > b$.

But think twice. You have to be sure your proof for the derivative of $x^2$ doesn't somehow rely on algebra equivalent to what (for some reason) you want to avoid.

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You can use the property of Arithmetic Mean(AM) and Geometric Mean (GM) Consider $2$ numbers $x, y \ge 0$ then $$\frac{x+y}{2} \ge \sqrt{xy},$$ Taking $x$ and y as $a^{2}$ and $b^{2}$ respectively, we can derive the desired result.

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Let $f(x)=e^{x}$, then $f''(x)=e^{x}>0$, $\forall x\in R$. Consequentyl, due to positive definite curvature, by Jensen's inequality it follows that for two $x,y \in R$ $$\frac{e^{x}+e^{y}}{2} \ge e^{\frac{x+y}{2}} \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}, ~~ \mbox{as we set}~ e^x=a~ and~ e^y=b, \mbox{where}~ a,b >0 $$

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Nothing changes if we assume that $a, b \in \Bbb R$ and $(b - a) \gt 0$.

$\quad a^2 + b^2 \gt 2ab \text{ iff }$
$\quad \quad b^2 -ab > ab - a^2 \text{ iff }$
$\quad\quad b (b - a) \gt a (b-a) \text{ iff }$
$\quad\quad b \gt a$
$\quad\quad (b -a) \gt 0$

When looking this over I can't help but think of an alliterative phrase:

$\quad$tautological truth

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