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Let $x,y,z \in \mathbb R^+$ such that $xyz\cdot(x+y+z) = 1$

Find $\min\{(x+y)(x+z)\}.$

Using calculus, and Lagrange multipliers, I get:

$(x+y)(x+z) \ge2$ (with the equality occurring if and only if $y=z=1,\ x=\sqrt{2} - 1$).

But I want to solve it in an easy way, which doesn't need calculus.

How can I do it? Thanks.

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Hint: Your expression is $x(x+y+z)+yz= 1/yz+yz$

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Hint: Recall the Heron's formula for triangle. Let $y+z=a, z+x=b, x+y=c$, then the problem becomes given the area to minimize $bc$.

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By AM-GM $$(x+y)(x+z)=x(x+y+z)+yz\geq2\sqrt{xyz(x+y+z)}=2.$$ The equality occurs for $xyz(x+y+z)=1$ and $x(x+y+z)=yz$,

which says that the equality indeed occurs, which gives the answer: $2$.

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