Showing $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$?

Question

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).

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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 $x,y \in R$ $$\frac{e^{x}+e^{y}}{2} \ge e^{\frac{x+y}{2}} \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}, ~~ \mbox{as we set}~ e^x=a~ and~ e^y=b, \mbox{where}~ a,b >0 $$

Answer 5

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. $\square$

Answer 6

If $a = b$ then $a^2 + b^2 = 2ab$.

Also, we can write as true

$\quad a^2 + b^2 \gt 2ab \; \text{ iff }$
$\quad \quad b^2 -ab > ab - a^2 \;\text{ iff }$
$\quad\quad b (b - a) \gt a (b-a) \;\text{ iff }$
$\quad\quad b \gt a \lor b \lt a \;\text{ iff }$
$\quad\quad a \ne b$

Answer 7

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.