Nesbitt's Inequality and I <Valentine Special>

Feb 15, 2026

For some unknown reason, I really really like Nesbitt’s inequality, and I want to share (and know) why.

What Is It?

Nesbitt’s inequality states that for all $a,b,c>0$ ,

\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}.

This is actually pretty interesting if you think about it. No matter what positive numbers you choose, how different they are, big small, medium, the equation will alwasy be greater than or equal $\frac{3}{2}$ .

First Look

First time that I saw this, I thought something wrong with it, like surely you can break it somehow.

So I tested it mentally with some numbers.

If $a=b=c$ , then each term is $\frac{a}{a+a}=\frac12$ , so the sum is $\frac32$ . Okay.
If $a=100$ , $b=c=1$ , then

\frac{100}{1+1}+\frac{1}{1+100}+\frac{1}{100+1} =50+\frac{1}{101}+\frac{1}{101} >50.

Still $\ge \frac32$ . Not even close.

At this point it stops being “maybe it’s false” and turns into “why does it refuse to go below $\frac32$ ?”
That’s the part I love: it feels like there’s some invisible floor.

The Tiny Secret: Each Term is “Half Plus Something”

Here’s a little rewrite that makes Nesbitt feel less like black magic.

Notice that

\frac{a}{b+c} =\frac{a}{(a+b+c)-a}.

Let $s=a+b+c$ . Then each term is $\frac{a}{s-a}$ . If $a$ is big, then $s-a$ is small, so that term explodes. If $a$ is small, then $s-a$ is basically $s$ , so the term shrinks.

So the terms “trade off” with each other: one getting huge forces the others to be small, but not too small in total. There’s always enough leftover mass in the cycle to keep the sum from dipping.

Proof

This is the cleanest proof I know, and it’s basically one inequality wearing a trench coat.

We use Cauchy–Schwarz:

\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3} \ge \frac{(x_1+x_2+x_3)^2}{y_1+y_2+y_3} \quad\text{for } y_i>0.

Now set

x_1=a,\;x_2=b,\;x_3=c, \qquad y_1=a(b+c),\;y_2=b(c+a),\;y_3=c(a+b).

Then

\frac{x_1^2}{y_1}=\frac{a^2}{a(b+c)}=\frac{a}{b+c},

and similarly for the others. So the left-hand side becomes exactly our Nesbitt expression:

\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.

Now simplify the denominator:

a(b+c)+b(c+a)+c(a+b) =ab+ac+bc+ab+ac+bc =2(ab+bc+ca).

So we have

\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)}.

Now the only thing left is to show

\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge \frac32.

Cross-multiply (everything is positive so it’s safe):

2(a+b+c)^2 \ge 3\cdot 2(ab+bc+ca) \quad\Longleftrightarrow\quad (a+b+c)^2 \ge 3(ab+bc+ca).

Expand the left:

a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca).

\Longleftrightarrow\quad a^2+b^2+c^2 \ge ab+bc+ca.

And that last inequality is always true, because

a^2+b^2+c^2-ab-bc-ca =\frac12\big((a-b)^2+(b-c)^2+(c-a)^2\big)\ge 0.

And we’re done:

\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac32.

When Do We Get Equality?

Equality happens exactly when $a=b=c$ .

You can kind of feel it:

If one of them is bigger, one term jumps up.
If one is smaller, the other terms “cover” for it.
The only way to sit exactly on the floor $\frac32$ is to make everything perfectly balanced.

So the minimum is not some weird corner case. It’s the most symmetric case possible.

Why I Think I Like It

Nesbitt’s inequality is one of those results that looks like it’s about fractions, but it’s really about balance.

It says: no matter how you distribute “weight” among $a,b,c$ , the cyclic structure refuses to let the total collapse. Symmetry is the minimum, chaos only pushes the sum upward.

And I don’t know, there’s something comforting about an inequality that’s basically saying:

you can try to ruin it, but it only gets bigger.