$\sqrt{a^2 +(1-a)^2}+ \sqrt{b^2 +(1-b)^2} \geq \sqrt{2} $

Question

$a,b \in \mathbb{R}$

My attempt:

If i can prove that: $$x^2 +(1-x)^2 \geq \frac{1}{2}$$ So therefore: $$\sqrt{x^2 +(1-x)^2} \geq \frac{\sqrt{2}}{2} $$ But i can’t prove the first inequality .

Answer 1

Why not, square it and put everything on LHS, get rid of the denumerator. You get $$(2x-1)^2\geq 0$$ which is true.

Answer 2

Let's see the minimum value of $\sqrt{x^2+(1-x)^2}$ $$=\sqrt{x^2+x^2-2x+1}$$ $$=\sqrt{2x^2-2x+1}$$ This expression will be minimum if the thing inside the root is minimum.

Thus, $$y=2x^2-2x+1$$ $$0=4x-2$$ $\text{"Differentiating both sides"}$ $$\implies x=\frac{1}{2}$$ Thus the function is minimum at $x=\frac{1}{2}$

Thus, minimum value of function is:$$2 \cdot(\frac{1}{2})^2-2 \cdot\frac{1}{2}+1$$ $$=\frac{1}{2}$$ Hence the minimum value of the expression $\sqrt{x^2+(1-x)^2}$ is $\sqrt{\frac{1}{2}}$ $=\frac{1}{\sqrt{2}}$

Thus the minimum value of $$\sqrt{a^2 +(1-a)^2}+ \sqrt{b^2 +(1-b)^2}$$ is : $$2 \cdot\frac{1}{\sqrt{2}}$$ $$=\sqrt{2}$$

Thus the required inequality has been proved.

Answer 3

Using the Cauchy-Schwarz inequality: $$ \sqrt{1^2 + 1^2} \cdot \sqrt{x^2 + (1-x)^2} \ge 1\cdot x + 1 \cdot (1-x) = 1 \, . $$

The geometric interpretation is that a point on the line $y=1-x$ in the 2D plane has at least the distance $1/\sqrt 2$ to the origin.

Yet another option is to substitute $x= \frac 12 + u$. Then $$ x^2 + (1-x)^2 = \left( \frac 12 + u\right)^2 + \left( \frac 12 - u\right)^2 = \frac 12 + 2 u^2 \ge \frac 12 \, . $$