gobathegreat wrote: Find minimum of $x+y+z$ where $x$, $y$ and $z$ are real numbers such that $x \geq 4$, $y \geq 5$, $z \geq 6$ and $x^2+y^2+z^2 \geq 90$ See also here https://artofproblemsolving.com/community/c6h457188p2568837 https://artofproblemsolving.com/community/c4h399079p2221319 Define $a=x^2, \; b=y^2 \; c=z^2$. Now we need to find min of $\sqrt{a}+\sqrt{b}+\sqrt{c}$ with $a\ge 16, \; b\ge 25, \; c\ge 36$ and $a+b+c\ge 90$. Using following lema (which is easy to prove): $x\le \min\{p,q \} \Longrightarrow \sqrt{p}+\sqrt{q} \ge \sqrt{p+q-x}+\sqrt{x}$, we have: $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge \sqrt{a}+5+\sqrt{b+c-25} \ge 4+5+\sqrt{a+b+c-25-16} \ge 16$. With equality when $(x,y,z)=(4,5,7)$. $\blacksquare$ https://artofproblemsolving.com/community/c6h604572p3590489 https://artofproblemsolving.com/community/c4h1144309p5389036

gobathegreat wrote: Find minimum of $x+y+z$ where $x$, $y$ and $z$ are real numbers such that $x \geq 4$, $y \geq 5$, $z \geq 6$ and $x^2+y^2+z^2 \geq 90$ Solution of Zhangyunhua:

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Another Solution:(sqing's method) $$\sum_{cyc} x = \sqrt{ \sum_{cyc} x^2 +2\sum_{cyc}xy} \geq \sqrt{90 +2xy+2(x+y)z} \geq \sqrt{90 +40 +2(x+y)\sqrt{90-(x^2+y^2)}} \geq \sqrt{90+40+126} =\boxed{16}$$