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If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$

Find the minimum value of the function

I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?

This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)

Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$

then

$A+B \geq 2\sqrt{AB}$

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

With equality if and only $A=B$ so $x=-\frac{1}{2}$

bigfocalchord
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    df(x)/dx=0 when y is minimum – Ariana Jul 28 '16 at 04:01
  • Sorry, but $x=-\frac12$ is not the correct answer.

    In fact, you first derived an inequality that says $f(x)\geq g(x)$, then you found $x^$ such that $f(x^)=g(x^)$, and finally concluded $f(x^)$ is the minimum of $f(x)$. Generally, this is not always a correct approach. For example, consider $f(x)=x^2$ and $g(x)=2x-1$. Starting from $(x-1)^2\geq0$, you can easily verify $f(x)\geq g(x)$. In this example, $f(x^)=g(x^)$ leads to ${x^}^2=2x^-1$, and then $x^*=1$. But obviously you can say $f(1)$ is the minimum of $f(x)$, because $f(0)=0$ is the true minimum.

     – babakks Jul 28 '16 at 07:00
  • But, if you find a constant function for $g(x)$ (i.e., $g(x)=k$), then you can say the minimum of $f(x)$ is $f(x^)$, if such $x^$ exists. In the aforementioned example, $g(x)=0$ is the constant function we need. So the correct answer of the problem is that explained by juantheron. – babakks Jul 28 '16 at 07:12

4 Answers4

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enter image description here

Let $A=(0,1)$, $B=(1,1)$, and $C=(-x,x)$ as in the picture. Then $$AC=\sqrt{x^2+(1-x)^2}, BC=\sqrt{(1+x)^2+(1-x)^2}.$$ Let $D$ be symmetric to $A$ about $x+y=0$ (trajectory of $C$). Apparently $$AC+BC\ge BD=AE+BE.$$ Minimal is attained at $C=E$. I leave you figure the coordinates of $E$.

Quang Hoang
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Using Minkowski Inequality

$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$$

and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$

So $$\sqrt{x^2+(1-x)^2}+\sqrt{(1-x)^2+(1+x)^2}\geq \sqrt{[x+(1-x)]^2+[(1-x)+(1+x)]^2}=\sqrt{5}$$

and Equality hold when $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-2x+1=x^2+x\Rightarrow x=\frac{1}{3}$$

juantheron
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A posible way: Let $A,B,C$ a triangle and $AD$ the height. The problem is $AD=1-x$, $BD=x$, $CD=1+x$ and you want minimize $BA+AC$, but note that the area is fixed ($(1-x)(2x+1)/2$), then this sum is minime if the triangle is right in A.

Martín Vacas Vignolo
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Write $f(x)=g(x)+h(x)$. Now $g(x)$ has minima at $x=1/2$ and $h(x)$ is increasing function with minima at $x=0$.Then $f(x)$ has minima at ?

Nitin Uniyal
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