Using geometry to prove an inequality

The inequality #

I really like this inequality, which I encountered in a past life when I used to do math competitions. The problem statement is as follows:

Problem. Prove that for all $a,b,c>0$ we have $$ \sqrt{a^2 - ab + b^2} + \sqrt{b^2 - bc + c^2} \ge \sqrt{a^2 + ac + c^2}. $$

Proof. Upon first glance this seems like a problem that requires a bunch of algebra to prove. But it turns out you can solve it by drawing a picture:

Draw a picture

In convex quadrilateral $ABCD$ above, let $BA=a$, $BD=b$, and $BC=c$ where $\angle ABD=\angle CBD=60^\circ$. By the Law of Cosines, $$ \begin{aligned} AD &= \sqrt{a^2-2ab\cos 60^\circ+b^2} = \sqrt{a^2-ab+b^2},\\ DC &= \sqrt{b^2-2bc\cos 60^\circ+c^2} = \sqrt{b^2-bc+c^2},\\ AC &= \sqrt{a^2-2ac\cos 120^\circ+c^2} = \sqrt{a^2+ac+c^2}. \end{aligned} $$

By the triangle inequality, we have $AD+DC\ge AC$, from which the conclusion follows.

An algebraic proof from GPT-5 #

I asked GPT-5 to prove this inequality algebraically with no geometric argument. It came up with the following proof:

Set $$ A=a^2-ab+b^2,\qquad B=b^2-bc+c^2,\qquad C=a^2+ac+c^2. $$ We want $\sqrt A+\sqrt B\ge \sqrt C$. Since both sides are nonnegative, it suffices to square once: $$ A+B+2\sqrt{AB}\ \ge\ C \iff 2\sqrt{AB}\ \ge\ C-(A+B). $$ A direct computation gives $$ C-(A+B)=ab+bc+ac-2b^2. $$ Therefore it remains to prove $$ 2\sqrt{AB}\ \ge\ ab+bc+ac-2b^2. $$ Observe that $$ \begin{aligned} \bigl(2\sqrt{AB}\bigr)^2-\bigl(ab+bc+ac-2b^2\bigr)^2 &=4AB-\bigl(ab+bc+ac-2b^2\bigr)^2\\[2mm] &=3\,(ab-ac+bc)^2\ \ge\ 0. \end{aligned} $$ Hence $2\sqrt{AB}\ge \bigl|ab+bc+ac-2b^2\bigr|\ge ab+bc+ac-2b^2$, which implies $A+B+2\sqrt{AB}\ge C$. Consequently, $$ \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge \sqrt{a^2+ac+c^2}. $$ Equality holds iff $ab-ac+bc=0$, i.e. $b=\dfrac{ac}{a+c}$.

Interactive visualization #

Finally, I asked GPT-5 to make an interactive app to visualize the inequality. The model is able to create a pretty cool visualization!

Click here for a fullscreen version.

Thanks to Andy Chen for reviewing a draft of this post.