Prove this inequality $\sum_{cyc}\frac{1}{x+3y}\ge\sum_{cyc} \frac{1}{z+2x+y}$

Question

Given $x,y,z$ are positive number. Prove that $$\frac{1}{x+3y}+\frac{1}{y+3z}+\frac{1}{z+3x}\ge \frac{1}{x+2y+z}+\frac{1}{y+2z+x}+\frac{1}{z+2x+y}$$

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we must prove $\frac{1}{x+2y+z}+\frac{1}{y+2z+x}+\frac{1}{z+2x+y}\le\frac{1}{4}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} )$ (Use $\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$)

And $\frac{1}{x+3y}+\frac{1}{y+3z}+\frac{1}{z+3x}\ge \frac{1}{4}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} )$ - How prove this inequality ?

Answer 1

By C-S $$\sum_{cyc}\frac{1}{2x+y+z}=7\sum_{cyc}\frac{1}{2(x+3y)+(y+3z)+4(z+3x)}=$$ $$=\frac{1}{7}\sum_{cyc}\frac{(2+1+4)^2}{2(x+3y)+(y+3z)+4(z+3x)}\leq$$ $$\leq\frac{1}{7}\sum_{cyc}\left(\frac{2^2}{2(x+3y)}+\frac{1^2}{y+3z}+\frac{4^2}{4(z+3x)}\right)=$$ $$=\frac{1}{7}\sum_{cyc}\left(\frac{2}{x+3y}+\frac{1}{y+3z}+\frac{4}{z+3x}\right)=\frac{1}{7}\sum_{cyc}\frac{7}{x+3y}=\sum_{cyc}\frac{1}{x+3y}$$ and we are done!

Answer 2

It is clear that the following expression is positive \begin{eqnarray*} 3\left(x(x^2-yz)^2+y(y^2-zx)^2+z(z^2-xy)^2\right)+ 1\left(yz^2(z-x)^2+zx^2(x-y)^2+xy^2(y-z)^2\right)+ 16\left(z(y^2-zx)^2+x(z^2-xy)^2+y(x^2-yz)^2\right)+ 2\left(xz^2(x-y)^2+zy^2(z-x)^2+yx^2(y-z)^2\right)+ 4\left(x^2z(z-y)^2+z^2y(y-x)^2+y^2x(x-z)^2\right) \end{eqnarray*} Multiply by this by two & do some algebra ... & the result follows.

((y+3z)*(z+3*x)*(x+2*y+z)*(y+2*z+x)*(z+2*x+y)+(x+3*y)*(z+3*x)*(x+2*y+z)*(y+2*z+x)*(z+2*x+y)+(x+3*y)*(y+3z)*(x+2*y+z)*(y+2*z+x)*(z+2*x+y)-(x+3*y)*(y+3z)*(z+3*x)*(y+2*z+x)*(z+2*x+y)-(x+3*y)*(y+3z)*(z+3*x)*(x+2*y+z)*(z+2*x+y)-(x+3*y)*(y+3z)*(z+3*x)*(x+2*y+z)*(y+2*z+x))-2*(3*(x*(x^2-y*z)^2+y*(y^2-z*x)^2+z*(z^2-x*y)^2)+1*(y*z^2*(z-x)^2+z*x^2*(x-y)^2+x*y^2*(y-z)^2)+16*(z*(y^2-z*x)^2+x*(z^2-x*y)^2+y*(x^2-y*z)^2)+2*(x*z^2*(x-y)^2+z*y^2*(z-x)^2+y*x^2*(y-z)^2)+4*(x^2*z*(z-y)^2+z^2*y*(y-x)^2+y^2*x*(x-z)^2));

The above statement can be verified by copying & pasting the above in reduce.

Answer 3

Let $a$, $b$ and $c$ be positive numbers.

Hence, by AM-GM $$\sum_{cyc}ab^3=\frac{1}{7}\sum_{cyc}(4ca^3+bc^3+2ab^3)\geq\frac{1}{7}\sum_{cyc}7\sqrt[7]{a^{14}b^7c^7}=\sum_{cyc}a^2bc.$$ Let $a=t^{x}$, $b=t^{y}$ and $c=t^{z}$, where $t>0$.

Hence, $$\sum_{cyc}\left(t^{x+3y}-t^{2x+y+z}\right)\geq0$$ or $$\sum_{cyc}\left(t^{x+3y-1}-t^{2x+y+z-1}\right)\geq0.$$ Thus, $$\int\limits_{0}^1\sum_{cyc}\left(t^{x+3y-1}-t^{2x+y+z-1}\right)dt\geq0$$ or $$\sum_{cyc}\left(\frac{1}{x+3y}-\frac{1}{2x+y+z}\right)\geq0$$ and we are done!

Answer 4

There is also the following ugly proof.

After full expanding we need to prove that $$\sum_{cyc}(3x^5+16x^4y+x^4z+20x^3y^2+2x^3z^2-8x^3yz-34x^2y^2z)\geq0,$$ which is obvious by Muirhead, AM-GM and Rearrangement.