# Find the minimum of $\frac{a^2+b^2}{2}-\frac{|a-b|}{2}\sqrt{1+(a+b)^2}$ [closed]

When $$a\neq b$$ $$a,b\in\mathbb{R}$$ find the minimum of $$\frac{a^2+b^2}{2}-\frac{|a-b|}{2}\sqrt{1+(a+b)^2}$$

I think that the answer is $$-\frac{1}{4}$$ but not sure. Please tell me a solution.

• How did you arrive at your answer? Apr 30, 2022 at 22:57
• Have you tried differentiating the function? Apr 30, 2022 at 22:58
• I thought that the answer will be when a=-b and then it could be written as $a^2-a$ from this the minimum should be $-\frac{1}{4}$ when $a=\frac{1}{2}$
– user998872
Apr 30, 2022 at 23:28
• I could not prove that it should be $a=-b$
– user998872
Apr 30, 2022 at 23:29
• Using $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$, we have $$\frac{a^2+b^2}{2}-\frac{|a-b|}{2}\sqrt{1+(a+b)^2} = \left(\frac{|a - b|}{2} - \frac 12\sqrt{1 + (a + b)^2}\right)^2 - \frac14.$$ May 1, 2022 at 6:05

Since swapping $$a$$ and $$b$$ does not change the value of the expression, let's restrict ourselves only to solutions where $$a\geq b$$. This allows us to rewrite $$|a-b|$$ as $$a-b$$, which makes our expression much easier to deal with.
Let $$c=\frac{a+b}{2}$$ and $$d=\frac{a-b}{2}$$. Then $$c$$ can be any real number, and $$d$$ can be any non-negative real. Written in terms of $$c$$ and $$d$$, our expression becomes:
$$c^2 + d^2 - d\sqrt{4c^2+1}$$
Differentiating with respect to $$d$$ gives $$2d - \sqrt{4c^2+1}$$. This is zero when $$d = \sqrt{c^2 + 1/4}$$ which means this value of $$d$$ minimises the expression.
If you plug this value of $$d$$ into the expression, the $$c$$s cancel out and the expression reaches its minimum value of $$-1/4$$ as you suggested.
If you have a particular value for $$a$$ and you want a corressponding value of $$b$$ that achieves this minimum, remember that $$a=c+d$$ and $$d = \sqrt{c^2 + 1/4}$$. Solving this gives $$c=\frac{a}{2}-\frac{1}{8a}$$ and $$d=\frac{a}{2}+\frac{1}{8a}$$. We get $$b=c-d =-\frac{1}{4a}$$. This only works if this value of $$b$$ is less than or equal to $$a$$, which we see is only when $$a>0$$.