Suppose we need to prove $$\forall x\in \mathbb{R} :f(x)=\sqrt{x^2-2\sqrt2x+9}+\sqrt{x^2-4\sqrt2x+16}\geq 5$$
First I use plotting to show this and find $min$ of the function is greater than $5$
Then I used derivation anf find $f'=0$ root in $x\sim 2.15 $ ,by tracing sign of $f'$ we can easily show that point is $min$
$$f'=0 \to x\sim 2.098 \implies min=(2.098,5.65)$$ so $f(x)\geq 5$ .
The original question was about $k-10$ students , so we can't use derivation or plotting. My question is : How to prove it by an elementary method ?
My trial was :$$f(x)=\sqrt{(x-\sqrt2)^2+7}+\sqrt{(x-2\sqrt2)^2+8}$$This is obvious $$\sqrt{(x-\sqrt2)^2+7}+\sqrt{(x-2\sqrt2)^2+8}\geq \sqrt 7+\sqrt 8>5 $$ But :$\bf \text {at the same time ,it is impossible }(x-\sqrt2)^2=0,(x-2\sqrt2)^2=0$
How can I describe it to $k-10 $ students ?
I truly appreciate your time and effort (for any Idea)
You can just explain that if $g(x),h(x)>0$ then $$\min [g(x)+h(x)] \geq \min g(x) + \min h(x). $$