# How to find minimum value

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.

• What have you tried? Commented Jan 28, 2019 at 9:42
• I tried to differentiate it with respect to x Commented Jan 28, 2019 at 9:43
• I tried to differentiate it with respect to x Commented Jan 28, 2019 at 9:43
• Please typeset your equations using Mathjax for better presentation Commented Jan 28, 2019 at 9:43

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 $$x=\sqrt{3\cdot5-\frac{3}{8}\sqrt{34}\cdot\frac{5}{8}\sqrt{34}}=\frac{15}{4\sqrt2}.$$