$a^2 + b^2 + 25 = ab + 5a + 5b$
I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful
The equation does look similar to an equation of a circle - can you use this idea?
The intuition is that the right-hand side is, with a few possible exceptins, smaller than the left-hand side.
Note that $(a-b)^2\ge 0$, so $ab\le \frac{a^2+b^2}{2}$. Thus the right-hand side is $\le \frac{a^2+b^2}{2}+5a+5b$.
It follows that $$(a^2+b^2+25)-(ab+5a+5b)\ge \frac{a^2+b^2}{2}+25-5a-5b.$$ The right=hand side above is $$\frac{1}{2}\left(a^2+b^2-10a-10b+50\right).$$ This is $$\frac{1}{2}\left((a-5)^2+(b-5)^2\right),$$ which is $\gt 0$ unless $a=b=5$.
Remark: We can make the proof more myaterious by writing the magic identity $$2\left[(a^2+b^2+25)-(ab+5a+5b)\right]=(a-5)^2+(b-5)^2+(a-b)^2,$$ and concluding that the left-hand side is $0$ precisely if $a=b=5$.
Let $x = a+b \to x^2 - 2ab + 25 = ab + 5x \to x^2 -5x + 25 = 3ab \leq 3\cdot \dfrac{(a+b)^2}{4} = \dfrac{3x^2}{4} \to 4x^2 - 20x + 100 \leq 3x^2 \to x^2 - 20x + 100 \leq 0 \to (x-10)^2 \leq 0 \to x = 10, a = b = \dfrac{10}{2} = 5$.
