Question

[Ngân Bướm To]

Cho x,y,z > 0 và x+y+z=3

cm \(\frac{x}{x+\sqrt{3x+yz}}+\frac{y}{y+\sqrt{3y+zx}}+\frac{z}{z+\sqrt{3z+xy}}\le1\)

Answer 1

[Kiệt Nguyễn]

Ta có: \(\left(x-\sqrt{yz}\right)^2\ge0\Rightarrow x^2+yz\ge2x\sqrt{yz}\)(Dấu "="\(\Leftrightarrow x^2=yz\))

Theo đề: x + y + z = 3\(\Rightarrow3x+yz=\left(x+y+z\right)x+yz=x^2+yz+x\left(y+z\right)\)\(\ge x\left(y+z\right)+2x\sqrt{yz}\)

Suy ra \(\sqrt{3x+yz}\ge\sqrt{x\left(y+z\right)+2x\sqrt{yz}}=\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)

và \(x+\sqrt{3x+yz}\ge\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\frac{x}{x+\sqrt{3x+yz}}\le\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự ta có: \(\frac{y}{y+\sqrt{3y+zx}}\le\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\);\(\frac{z}{z+\sqrt{3z+xy}}\le\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng từng vế của các BĐT trên,ta được:

\(\frac{x}{x+\sqrt{3x+yz}}\)\(+\frac{y}{y+\sqrt{3y+zx}}\)\(+\frac{z}{z+\sqrt{3z+xy}}\le1\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

Answer 2

[Nyatmax]

We have:

\(VT=\Sigma_{cyc}\frac{x}{x+\sqrt{3x+yz}}=\Sigma_{cyc}\frac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}=\Sigma_{cyc}\frac{\frac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}}{\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+1}\)

Dat \(\left(\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}};\frac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}};\frac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\right)=\left(a;b;c\right)\)

Consider:

\(\Sigma_{cyc}\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\Sigma_{cyc}\frac{\frac{x}{x+y}+\frac{x}{x+z}}{2}=\frac{3}{2}\)

\(\Rightarrow a+b+c\le\frac{3}{2}\)

Now we need to prove:

\(\Sigma_{cyc}\frac{a}{a+1}\le1\)

\(\Leftrightarrow\Sigma_{cyc}\frac{1}{a+1}\ge2\left(M\right)\)

\(VT_M\ge\frac{9}{a+b+c+3}\ge\frac{9}{\frac{3}{2}+3}=2\)

Sign '=' happen when \(\hept{\begin{cases}x=y=z=1\\a=b=c=\frac{1}{2}\end{cases}}\)