Review question

# Can we prove that, for a triangle, $4(s-b)(s-c) \le a^2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8056

## Solution

The sides of a triangle are $a$, $b$, $c$, and $s$ is given by $2s = a + b + c$.

Prove that

1. $4(s-b)(s-c) \le a^2$,
2. $8(s-a)(s-b)(s-c) \le abc$.
For the first inequality, notice that \begin{align*} 4(s-b)(s-c) &= 4 \left( \frac{a+b+c}{2} - b \right) \left( \frac{a+b+c}{2} - c \right) \\ &= (a+b+c-2b)(a+b+c-2c) \\ &= (a+c-b)(a+b-c) \\ &= (a+(c-b))(a-(c-b)) \\ &= a^2 - (c-b)^2 \\ &\le a^2 \end{align*}

with equality if and only if $b = c$. This holds irrespective of the values of $a$, $b$, and $c$; they needn’t even be positive.

We’ll turn now to the second inequality. By exchanging the roles of $a$, $b$, and $c$ in the above, we also have the inequalities $\begin{equation*} 4(s-a)(s-c) \le b^2 \quad\text{and}\quad 4(s-a)(s-b) \le c^2. \end{equation*}$

As the numbers $a$, $b$ and $c$ are the side lengths of a triangle, we know that $a \le b + c$, $b \le a + c$, and $c \le a + b$.

Thus, $s - a \ge 0$, $s - b \ge 0$ and $s - c \ge 0$, and so the left-hand side of the three inequalities of type (i) are all non-negative.

We can therefore multiply the three inequalities together while preserving the inequality, which leads to

$\begin{equation*} 4^3(s-a)^2(s-b)^2(s-c)^2 \le a^2b^2c^2 \iff (8(s-a)(s-b)(s-c))^2 \le (abc)^2. \end{equation*}$

As all of the terms are non-negative, we can take square roots of both sides, which leads us to inequality (ii).

State under what conditions the sign of equality applies, in each case.

For inequality (i), the proof shows that equality holds exactly when $(c-b)^2 = 0$, that is, when $b = c$, so the triangle is isosceles.

For inequality (ii), we have equality when we have equality in all three versions of (i).

So here $a = b = c$, and the triangle is equilateral.

Investigate the truth of the results (i) and (ii) if $a$, $b$, $c$ are any positive quantities, not necessarily the sides of a triangle.

The inequality (i) holds for any values of $a$, $b$ and $c$, irrespective even of their sign.

For (ii), we can suppose without any loss of generality that $c \ge b \ge a > 0$.

Then $a < b + c$ and $b < a + c$, so that $s - a > 0$ and $s - b > 0$.

We cannot necessarily say that $c \le a + b$, or that $s - c \ge 0$. There are two cases to consider.

If $s - c \ge 0$, then the same argument that we employed earlier can be applied again, and the inequality holds. In this case, $a$, $b$ and $c$ form the sides of a triangle (which is degenerate if $c=a+b$).

If $s - c < 0$, then the left-hand side is the product of three terms, two of which are positive and one of which is negative, and so the left-hand side is negative while the right-hand side is positive; the inequality is therefore true in this case too.

Therefore, inequality (ii) is true for all positive values of $a$, $b$ and $c$.

This argument also works if we allow some or all of $a$, $b$ and $c$ to be zero.

However, if we allow them to be negative, then (ii) does not necessarily hold.

For example, take $a = b = 2$ and $c = -2$; in this case, the left-hand side of (ii) is positive while the right-hand side is negative.