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

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 $$0 but I don't think restricting $$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:

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

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).

• If $a>b$, then we do have $2a^2>2ab$ for your case of interest (where $a>0$). – Minus One-Twelfth May 18 at 20:12
• 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. – lulu May 18 at 20:14

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

• 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 $aa+ab > 2ab$? – user106860 May 18 at 20:53
• @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. – Mark Bennet May 18 at 21:48

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. $$\square$$

• This is cool. Thank you. – user106860 May 18 at 20:57

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.

• 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. – Billy Rubina May 19 at 3:26
• @BillyRubina Personally, I would stick to $\left(a-b\right)^{2}\geq 0$. – eranreches May 19 at 10:10

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.

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.

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

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