# Tagged Questions

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### minimal value of $x^2+2y^2+5z^2$ with constraint.

$x,y,z>0$, and $xy+yz+zx=1$. I need to find the minimum value of $x^2+2y^2+5z^2$ In general what can we say about the minimal value of $\frac{ax^2+by^2+cz^2}{xy+xz+yz}$, over all positive numbers ...
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### The expression $1 + x^2 +(-T_px+y)^2 +z^2$ is bounded below by a constant multiple of $(1+x^2+y^2+z^2)$

Suppose $T_p > 0$. Is there a simply way to show that $1 + x^2 +(-T_px+y)^2 +z^2 \geq C (1+x^2+y^2+z^2)$, for all $(x,y,z) \in \mathbb R^3$, where $C>0$.
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### verifying extrema found by Lagrange multipliers

This question was inspired by reading this problem: Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Suppose I have a function $f(x,y,z)$ with continuous ...
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### What is the domain of $z=\arcsin\dfrac{x}{y}$?

I get that it should be $|y|>|x|$ and in the Wolfram it looks like this. But when I graph it by hand is that it should be only the "upper" part of intersection and not the "bottom" part as well, ...
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### Calculus-based proof that $x_1^{p_1}\cdots x_n^{p_n}\le p_1x_1+\dots+p_nx_n$ when $\sum p_i=1$

Let $$g(x_1...x_n)=x_1^{p_1}\cdot...x_n^{p_n}$$ $$u(x_1...x_n)=p_1x_1+...p_nx_n$$ Where $\sum p_i = 1$. I have to show that $f(x)=g(x)-u(x)$ is always negative or $0$ over $\Bbb R_+^n$. I've ...
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### Show that $\forall (x,y)$ in the first quadrant: $\frac {x^2+y^2}{4}\leq e^{x+y-2}$

I have the folowing exercise (which I've been thinking quite a while and couldn't figure out): Show that $\forall (x,y)$ in the first quadrant: $$\frac {x^2+y^2}{4}\leq e^{x+y-2}$$ My idea was to ...
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### Maximization of an integer input function

Maximize the value of the function $$z=\frac{ab+c}{a+b+c},$$ where $a,b,c$ are natural numbers and are all lesser than 2010 and not necessarily distinct from each other. Please provide a proof, ...
Suppose that $\sum_{i=1}^{n}\lambda_{i}=1$, where $\lambda_{i}>0$, and $\sum_{i=1}^{n}x_{i}^{2}=1$, where $x_{i}>0$. Does one have $n^{3/2}\min_{1\le i\le n}\lambda_{i}x_{i}\le B$ for some ...