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I would appreciate if somebody could help me with the following problem.

Let $ x, y $ be reals such that $x^2+y^2=4 .$ Prove that$$42>2\sqrt{(x-5)^2+y^2} + 5\sqrt{x^2+(y-4)^2}\geq 4\sqrt{26}$$

My work : The answer can be found using differentiation by substituting $x=\cos t$, $y=\sin t$ in the given range, but I want to show that the inequality holds without using calculus.

Thanks!

Young
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    The minimum is exactly $4\sqrt{26}$ when $x = \frac{10-\sqrt{22}}{13}$ and $y = \frac{2 + 5\sqrt{22}}{13}$. – River Li Nov 11 '23 at 00:21

2 Answers2

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For upper bound:

With $x=2\cos\theta$ and $y=2\sin\theta$, $$d=2\sqrt{(x-5)^2+y^2} + 5\sqrt{x^2+(y-4)^2}=2\sqrt{29-20\cos\theta}+5\sqrt{20-16\sin\theta}.$$ By Jensen's inequality $$d\leq 7\sqrt{\frac27(29-2\cos\theta)+\frac57(20-16\sin\theta)}=\sqrt7\sqrt{158-40\sqrt5\cos(\theta-\alpha)}$$ where $\cos\alpha=\frac1{\sqrt5}$ and $\sin\alpha=\frac2{\sqrt5}$. Hence $$d\leq\sqrt7\sqrt{158+40\sqrt5}<42.$$

Bob Dobbs
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COMMENT.-Clearly the inequality is equivalent to $$4\sqrt{26}\le2\sqrt{29-10x}+5\sqrt{20-8y}\lt42$$ We can looking for the extremums of the function $F(x,y)=2\sqrt{29-10x}+5\sqrt{20-8y}$ conditionated by $x^2+y^2=4$ and if we prefer to work with one variable, we have to consider the function (two cases because of the sign $\pm$) $$f_{\pm}(x)=2\sqrt{29-10x}+5\sqrt{20\pm8\sqrt{4-x^2}}\space$$ The derivatives are $$f'_+(x)=\frac{-10x}{\sqrt{(4-x^2)(5-2\sqrt{4-x^2})}}-\frac{10}{\sqrt{29-10x}}\\f'_{-}(x)=\frac{10x}{\sqrt{(4-x^2)(5-2\sqrt{4-x^2})}}-\frac{10}{\sqrt{29-10x}}$$ Approximative calculations give $$28.361\le f_+(x)\le41.609\\4\sqrt{26}\approx20.396078\le f_{-}(x)\le36.361$$ which shows that always $$4\sqrt{26}\le f_ {\pm}(x)\lt42$$

Piquito
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