Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:

\(\dfrac{xy}{4}=\dfrac{yz}{6}=\dfrac{xz}{10}=\dfrac{xy+yz+xz}{4+6+10}=\dfrac{60}{20}=3\)

=>xy=12; yz=18; xz=30

=>xyz=căn(12*18*30)=36căn 5

=>\(z=3\sqrt{5};x=2\sqrt{5};y=\dfrac{6\sqrt{5}}{5}\)

Bài này phải tìm GTLN chứ nhỉ?!

đé* biết ok

\(Q=\frac{1}{\frac{x}{y}+\frac{z}{x}+1}+\frac{1}{\frac{y}{z}+\frac{x}{y}+1}+\frac{1}{\frac{z}{x}+\frac{y}{z}+1}\)

Đặt \(\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)

\(Q=\frac{1}{a^3+c^3+1}+\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}\)

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

\(\Rightarrow Q\le\frac{1}{ac\left(a+c\right)+1}+\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}\)

\(Q\le\frac{abc}{ac\left(a+c\right)+abc}+\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}\)

\(Q\le\frac{b}{a+b+c}+\frac{c}{a+b+c}+\frac{a}{a+b+c}=1\)

\(\Rightarrow Q_{max}=1\) khi \(a=b=c=1\) hay \(x=y=z\)

Quẩy lên các em êii

\(A=\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\)

\(2A=\frac{z+2\sqrt{xy}}{z+2\sqrt{xy}}-\frac{z}{z+2\sqrt{xy}}+\frac{x+2\sqrt{yz}}{x+2\sqrt{yz}}-\frac{x}{x+2\sqrt{yz}}+\frac{y+2\sqrt{zx}}{y+2\sqrt{zx}}-\frac{y}{y+2\sqrt{zx}}\)

\(=3-\left(\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{zx}}+\frac{z}{z+2\sqrt{xy}}\right)\le3-\left(\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\right)\)

\(=3-\frac{x+y+z}{x+y+z}=3-1=2\)\(\Leftrightarrow\)\(A\le\frac{2}{2}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)

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