Matematické Fórum

stuart clark · 06. 05. 2019 09:12

If $x^2+y^2+z^2=1.$ Then maximum value of $(3\sqrt{2}y-\sqrt{11}z)^2+(\sqrt{7}z-3\sqrt{2}x)^2+(\sqrt{11}x-\sqrt{7}y)^2$ is

stuart clark

Thanks friends Got it.

Let $\vec{a} = x\hat{i}+y\hat{j}+z\hat{k}$ and $\vec{b} = \sqrt{7}\hat{i}+\sqrt{11}\hat{j}+3\sqrt{2}\hat{k}$

Now Using $\bigg|\vec{a}\times \vec{b}\bigg|^2=|\vec{a}|^2|\vec{b}|^2-\bigg(\vec{a}\cdot \vec{b}\bigg)\leq |\vec{a}|^2|\vec{b}|^2$

So

$(3\sqrt{2}y-\sqrt{11}z)^2+(\sqrt{7}z-3\sqrt{2}x)^2+(\sqrt{11}x-\sqrt{7}y)^2\leq 36$

Equality hold when $\sqrt{7}x+\sqrt{11}y+3\sqrt{z}=0.$

stuart clark

I have a doubt on that problem

Show that for any positive integer $n>11$

$\bigg(\frac{2n-e}{e}\bigg)^{\frac{2n-1}{2}}<1\cdot 3\cdot 5 \cdot 7 \cdots (2n-1)<\bigg(\frac{2n+e}{e}\bigg)^{\frac{2n+1}{2}}$

Plesse have a look on that problem. Thanks

Matematické Fórum

#1 06. 05. 2019 09:12

Maximum value

#2 08. 05. 2019 12:21 — Editoval stuart clark (09. 05. 2019 17:11)

Re: Maximum value

#3 08. 05. 2019 12:33 — Editoval stuart clark (08. 05. 2019 12:55)

Re: Maximum value

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