# Tagged Questions

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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 ...
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### $x^2+y^2=z^2(1+xy)$ prove $z=\min \{x;y;z\}$ (with $x,y,z \in \mathbb{Z^+}$)
$x,y,z \in \mathbb{Z^+}$ such that $x^2+y^2=z^2(1+xy)$. Prove $z=\min \{x;y;z\}$ $$x^2+y^2=z^2(1+xy) \iff xy = \frac{x^2+y^2} {z^2} - 1$$. Assum $z>y \implies xy < x^2/z^2$, we have \$xy \in Z ...