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Source: Bulgaria Spring Mathematical Competition 2025 12.1

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#1 h Mar 30, 2025, 1:15 PM

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In terms of the real numbers $a$ and $b$ determine the minimum value of $\sqrt{(x+a)^2+1}+\sqrt{(x+1-a)^2+1}+\sqrt{(x+b)^2+1}+\sqrt{(x+1-b)^2+1}$ as well as all values of $x$ which attain it.

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#2 h Mar 30, 2025, 1:51 PM

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Use Minkowski's Inequality.
$\sqrt{(x+a)^2+1}+\sqrt{(a-1-x)^2+1}\geq \sqrt{(2a-1)^2+4}$ $\sqrt{(x+b)^2+1}+\sqrt{(b-1-x)^2+1}\geq \sqrt{(2b-1)^2+4}$ Thus
$LHS\geq \sqrt{(2a-1)^2+4}+\sqrt{(2b-1)^2+4}$ the equality holds at $(2x+1)(a+b+1)=0$ so for $x=-1/2$ .

This post has been edited 4 times. Last edited by ehuseyinyigit, Mar 30, 2025, 2:02 PM

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#3 h Mar 30, 2025, 1:56 PM

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We can see that $\sqrt{(x+a)^2+1}+\sqrt{(x+1-a)^2+1}+\sqrt{(x+b)^2+1}+\sqrt{(x+1-b)^2+1}$ is exactly the sum of $XA+XA_1+XB+XB_1$ , where $X$ is $(x,1)$ , $A=(-a,0)$ , $A_1=(a-1,0)$ , $B=(-b,0)$ , $B_1=(b-1,0)$ . We will prove that this minimum is then $\displaystyle X=(-\frac{1}{2},1)$ , meaning that the minimum is when $\displaystyle x=-\frac{1}{2}$ .

Let $A_2$ be the reflection of $A$ with respect to the line $y=1$ . Now from the triangle inequality $A_1A_2\le XA_1+XA_2=XA_1+XA$ whose minimum is when $X\in A_1A_2$ , or $\displaystyle x=-\frac{1}{2}$ . Analougasly we can see that the minimum $XB+XB_1$ is when $\displaystyle x=-\frac{1}{2}$ , so the minimum value is

$\sqrt{(2a-1)^2+4}+\sqrt{(2b-1)^2+4}$ .

This post has been edited 3 times. Last edited by p.lazarov06, Mar 30, 2025, 2:02 PM

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#4 h Mar 30, 2025, 2:02 PM

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Minimizing $\sqrt{(x+a)^2+1}+\sqrt{(x+1-a)^2+1}$ via derivative directly (and doing the same for the part with $b$ ) suffices.

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