Let $y+z=m, \ \ yz=n, \ \ x=\frac{1-n}{m+2n}$. If $m\ge 2$, then $4x+y+z=4x+m\ge 2$ If $m\le 2$, then $4x+y+z\ge 2 \ \ \iff \ \ \frac{4(1-n)}{m+2n}+m\ge 2 \ \ \iff \ \ (m^2-4n)+2(1-n)(2-m)\ge 0$ It is true because $m^2-4n=(y-z)^2\ge 0, \ \ x=\frac{1-n}{m+2n}\ge 0$

mudok wrote: Let $y+z=m, \ \ yz=n, \ \ x=\frac{1-n}{m+2n}$. If $m\ge 2$, then $4x+y+z=4x+m\ge 2$ If $m\le 2$, then $4x+y+z\ge 2 \ \ \iff \ \ \frac{4(1-n)}{m+2n}+m\ge 2 \ \ \iff \ \ (m^2-4n)+2(1-n)(2-m)\ge 0$ It is true because $m^2-4n=(y-z)^2\ge 0, \ \ x=\frac{1-n}{m+2n}\ge 0$ What is this method, sir?

Akatsuki1010 wrote: What is this method, sir? There is no something special in this proof.

Ankoganit wrote: Positive numbers $x, y, z$ satisfy the condition $$xy + yz + zx + 2xyz = 1.$$Prove that $4x + y + z \ge 2.$ A. Khrabrov See also here http://www.artofproblemsolving.com/community/c6h1180667p6752983

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luofangxiang wrote: why $\frac{b}{a+c}+\frac{c}{a+b} \ge \frac{b+c}{2}.\frac{4}{2a+b+c}$ ?

rearrangement

Ankoganit wrote: Positive numbers $x, y, z$ satisfy the condition $$xy + yz + zx + 2xyz = 1.$$Prove that $4x + y + z \ge 2.$ A. Khrabrov $\frac{4x}{y+z} + \frac{y}{z+x} +\frac{z}{x+y} \geq 2$ $LHS + 6 = (\frac{4x}{y+z}+4)+(\frac{y}{z+x}+1)+(\frac{z}{x+y}+1) $ $LHS + 6 = (x+y+z)(\frac{4}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}) $ $= \frac{1}{2}((x+y)+(y+z)+(z+x))(\frac{4}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}) \geq \frac{1}{2}(2+1+1)^2 $ $= 8$ then $LHS \geq 2$

How did you factor that? Shouldn't it be:$$LHS + 6 = (x+y+z)(\frac{4}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}) $$How did you get the first step?

luofangxiang wrote: why it can be written as like this can you explain me

why $\frac{b}{a+c}+\frac{c}{a+b} \ge \frac{b+c}{2}.\frac{4}{2a+b+c}$ rearrangement or titu's lema

Ankoganit wrote: Positive numbers $x, y, z$ satisfy the condition $$xy + yz + zx + 2xyz = 1.$$Prove that $4x + y + z \ge 2.$ A. Khrabrov Proof: $$1=xy + yz + zx + 2xyz\leq x(y+z)+\frac{(y+z)^2}{4}+\frac{x(y+z)^2}{2}\iff((2x+1)(y+z)-2)(y+z+2)\geq 0$$$$\implies 2x+1\geq \frac{2}{y+z},$$$$4x + y + z \geq \frac{4}{y+z}+y+z-2\geq 2.$$

Ankoganit wrote: Positive numbers $x, y, z$ satisfy the condition $$xy + yz + zx + 2xyz = 1.$$Prove that $4x + y + z \ge 2.$ A. Khrabrov https://artofproblemsolving.com/community/c6h1180667p6757698 https://artofproblemsolving.com/community/c6h1963771p13595710 https://artofproblemsolving.com/community/c6h1703592p10973642 Let $ a, b, c \ge 0 .$ Prove that $$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{4c}{a+b} \geq 2$$https://artofproblemsolving.com/community/c6h1643160p10359803

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