Prove that a triangle exists using the triangle inequality.

Question

I need help with the following question:

It is given that there is a triangle with sides of lengths a, b and c units.

Prove that there exists a triangle having sides of lengths $\sqrt{a(b+c-a)}, \sqrt{b(a+c-b)}$ and $\sqrt{c(a+b-c)}$ units.

Here's what I have so far:

$\sqrt{a(b+c-a)} + \sqrt{b(a+c-b)} \geq \sqrt{c(a+b-c)}$

$a(b+c-a) + 2\sqrt{ab(b+c-a)(a+c-b)} + b(a+c-b) \geq {c(a+b-c)}$

$2ab - a^2 - b^2 + 2\sqrt{ab(b+c-a)(a+c-b)} \geq -c^2$

$2\sqrt{ab(b+c-a)(a+c-b)} -(a-b)^2 \geq -c^2$

$-2\sqrt{ab(b+c-a)(a+c-b)} \leq c^2 -(a-b)^2$

$-2\sqrt{ab(b+c-a)(a+c-b)} \leq (b + c - a)(a + c - b)$

I thought to divide both sides by $\sqrt{(b+c-a)(a+c-b)}$ but either $\sqrt{b+c-a}$ or $\sqrt{a+c-b}$ might be $0$. I am unable to proceed from here.

Answer 1

Let $b+c-a=x$, $a+c-b=y$ and $a+b-c=z$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$\sqrt{(y+z)x}+\sqrt{(x+z)y}>\sqrt{(x+y)z},$$ which is true because $$\sqrt{(y+z)x}+\sqrt{(x+z)y}>\sqrt{(y+z)x+(x+z)y}>\sqrt{(x+y)z}.$$ By the same way we can prove another two inequalities.

Answer 2

Let us assume that there exists a triangle having sides of lengths $\sqrt{a(b+c-a)}, \sqrt{b(a+c-b)}$ and $\sqrt{c(a+b-c)}$ units.

$\implies$ $\sqrt{a(b+c-a)} + \sqrt{b(a+c-b)} \geq \sqrt{c(a+b-c)}$

Clearly both sides are positive, hence we may square both sides without disturbing the inquality,

$\implies$ $a(b+c-a) + 2\sqrt{ab(b+c-a)(a+c-b)} + b(a+c-b) \geq {c(a+b-c)}$

$2ab - a^2 - b^2 + 2\sqrt{ab(b+c-a)(a+c-b)} \geq -c^2$

$2\sqrt{ab(b+c-a)(a+c-b)} -(a-b)^2 \geq -c^2$

$-2\sqrt{ab(b+c-a)(a+c-b)} \leq c^2 -(a-b)^2$

$-2\sqrt{ab(b+c-a)(a+c-b)} \leq (b + c - a)(a + c - b)$

Now, RHS is neccesarily positive and for the case of LHS we got $2\sqrt{ab(b+c-a)(a+c-b)}$ positive too $\implies$ $-2\sqrt{ab(b+c-a)(a+c-b)}$ is negative.

Since, $f(x)=x^2$ is decreasing in the interval $x\le0$, we must flip the inequality,

$\implies$ By squaring both sides again, we get, $4{ab(b+c-a)(a+c-b)} \ge {[(b+c-a)(a+c-b)]}^2$

Dividing both sides by $(b+c-a)(a+c-b)$ , i.e. positive, we get, $4ab \ge (b+c-a)(a+c-b) =c^2 -(a-b)^2$

$(a-b)^2+4ab \ge c^2$

$(a+b)^2 \ge c^2$

Square-rooting both sides, we get, $ a+b \ge c$ which is true since it is given that there exists a triangle with sides of lengths a, b and c units.

Therefore, our assumption is true and there exists a triangle having sides of lengths $\sqrt{a(b+c-a)}, \sqrt{b(a+c-b)}$ and $\sqrt{c(a+b-c)}$ units.