# Tagged Questions

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### Find the smallest constant K satisfying the inequality

Find the smallest constant $K$satisfying the inequality $$x^{1\over 3}+y^{1\over 3} \le K(x+y)^{1\over 3}$$ The official proof makes the substitution $a=x^{1\over 3}$ and $b=y^{1\over 3}$, which does ...
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### How to show that $\frac {q + \frac {1}{2}}{p - \frac {1}{2}} > \sum_{i = p}^q\frac {1}{i}$ if $q\ge p > 0?$

How to show that : $$\frac{2q+1}{2p-1}>\sum_{i=p}^q\frac{1}{i}$$ if $q\ge p>0$
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### How prove that $q \geq b+d$ for $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$?

Let $a,b,c,d,p$, and $q$ be natural numbers such that $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$. How prove that $q \geq b+d$?
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### How prove that $x+y+z>4$ for $a+b+c=4$ and $ax+by+cz=xyz$?

Given positive reals $a,b,c,x,y,z$ such that $a+b+c=4$ and $ax+by+cz=xyz$. How prove that $x+y+z>4$?
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### Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers.

I stumbled upon it on some olympiad papers. Tried to AM>GM but didn't get any idea to move forward.
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### Arithmetic Mean and Geometric Mean Question, Guidance Needed

I am super new to olympiad-style math which focuses on a lot of inequalities, and tough problems which highschool students do not go over. I'm in 9th grade, and am trying to get into all of this stuff ...
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### A Cauchy-Schwartz type inequality

Given positive integers $k<n$ and positive real numbers $x_1$, $x_2, \dots, x_n$. Denote $$A={x_1\over x_2+x_3+\dots+x_{k+1}}+{x_2\over x_3+x_4+\dots+x_{k+2}}+\ldots+{x_n\over x_1+x_2+\dots+x_k}$$ ...
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### Proving a sharp inequality

After experimenting, I've come to the conclusion that if $x\geq y\geq z\geq 0:$ $$\sum_{x,y,z}\frac{x}{\sqrt{x+y}}\geq \sum_{x,y,z}\frac{y}{\sqrt{x+y}}$$ (the sums are cyclic) Does anyone know how ...
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If: $x^2+y^2+z^2=2(xy+xz+zy)$ and $x,y,z \in R^+$ Prove: $\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$ I tried my best to solve this thing but no use. Hope you guys can help me.Thanks in advance.
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### Find the maximum value of $abc$

$a,b,c$ are three positive real numbers such that $ab+bc+ca=12$. Then find the maximum value of $abc$
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### Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$

Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$. (Source : Singapore Math Olympiad ...
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### Finding the value of $(bc-ad)(ac-bd)(ab-cd)$

Let $a,b,c,d$ be $4$ distinct non-zero integers such that $a+b+c+d = 0$. It is know that the number $$M = (bc - ad)(ac - bd)(ab-cd)$$ lies strictly between $96100$ and $98000$. Determine the value ...
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### Given $x^2 + y^2 + z^2 = 3$ prove that $x/\sqrt{x^2+y+z} + y/\sqrt{y^2+x+z} + z/\sqrt{z^2+x+z} \le \sqrt3$

Given $x^2 + y^2 + z^2 = 3$ Then prove that $${x\over\sqrt{x^2+y+z}} + {y\over\sqrt{y^2+x+z}} + {z\over\sqrt{z^2+x+y}} \le \sqrt 3$$ I tried using Cauchy-Schwartz inequality but the inequality is ...
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### If $a,b,c,d>0$ and $a+b+c+d=4$, then $\frac{1}{11+a^2}+\frac{1}{11+b^2}+\frac{1}{11+c^2}+\frac{1}{11+d^2} \leq \frac {1}{3}$

Prove if $a,b,c,d>0$ and $a+b+c+d=4$, then $$\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \dfrac {1}{3}$$ This was an Inequality Olympiad Problem1. I proved by ...