How to find minimum value

Question

For $a=\sqrt{x^2-3\sqrt2x+9}$ and $b=\sqrt{x^2-5\sqrt2x+25}$ what is the value of $x$ when $a+b$ is minimum and how to find this? Thanks in advance.

Answer 1

Let $\measuredangle ACB=90^{\circ},$ $AC=3$, $BC=5$ and $CD$ be a bisector of $\angle ACB$.

Also, let $CD=x$.

Thus, by the triangle inequality $$AD+BD\geq AB,$$ which gives $$a+b=\sqrt{x^2+3^2-2x\cdot3\cdot\cos45^{\circ}}+\sqrt{x^2+5^2-2x\cdot5\cdot\cos45^{\circ}}\geq\sqrt{3^2+5^2}=\sqrt{34}.$$ The equality occurs, when $D\in AB$, which says that we got a minimal value.

Now, by similarity we can show that $$CD^2=AC\cdot BC-AD\cdot BD$$ and since $$\frac{AD}{BD}=\frac{AC}{BC}=\frac{3}{5},$$ we obtain $$AD=\frac{3}{8}\sqrt{34},$$ $$BD=\frac{5}{8}\sqrt{34}$$ and $$x=\sqrt{3\cdot5-\frac{3}{8}\sqrt{34}\cdot\frac{5}{8}\sqrt{34}}=\frac{15}{4\sqrt2}.$$