Đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\Rightarrow a^3b^3+b^3c^3+c^3a^3=1\)

\(=\sum\dfrac{a^{12}}{a^6+b^6}=\sum\dfrac{a^6\left(a^6+b^6\right)}{a^6+b^6}-\sum\dfrac{a^6b^6}{a^6+b^6}\\ =\sum a^6-\sum\dfrac{a^6b^6}{a^6+b^6}\\ \overset{Cosi}{\ge}a^3b^3+b^3c^3+c^3a^2-\sum\dfrac{a^6b^6}{2a^3b^3}\\ =1-\dfrac{1}{2}\sum a^3b^3=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Dấu = xảy ra khi \(x=y=z=\dfrac{1}{\sqrt[3]{3}}\)

\(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3\)

\(\Rightarrow3\ge3\sqrt[3]{\left(ab.bc.ca\right)^3}=3\left(abc\right)^2\Rightarrow a^2b^2c^2\le1\)

Ta có: \(\dfrac{a^{10}}{b^2c^2}+a^2b^2c^2\ge2a^6\)

Tương tự và cộng lại: \(P+3\left(abc\right)^2\ge2\left(a^6+b^6+c^6\right)\)

\(\Rightarrow P\ge2\left(a^6+b^6+c^6\right)-3a^2b^2c^2\ge2\left[\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3\right]-3=3\)

Lời giải:

Áp dụng BĐT AM-GM:
$\frac{x^3}{y(x+z)}+\frac{y}{2}+\frac{x+z}{4}\geq \frac{3}{2}x$

Tương tự với các phân thức còn lại, cộng theo vế và rút gọn ta được:

$\Rightarrow P=\sum \frac{x^3}{y(x+z)}\geq \frac{x+y+z}{2}$

Tiếp tục áp dụng AM-GM:

$x+y\geq 2\sqrt{xy}$

$y+z\geq 2\sqrt{yz}$

$x+z\geq 2\sqrt{xz}$

$\Rightarrow x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1$

$\Rightarrow P\geq \frac{1}{2}$

Vậy $P_{\min}=\frac{1}{2}$ khi $x=y=z=\frac{1}{3}$

\(\dfrac{x^3}{y\left(x+z\right)}+\dfrac{y}{2}+\dfrac{x+z}{4}\ge\dfrac{3x}{2}\)

Tương tự và cộng lại:

\(P+x+y+z\ge\dfrac{3}{2}\left(x+y+z\right)\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(x+y+z\right)\ge\dfrac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\dfrac{1}{2}\)

Ta có :

\(P=\sum\dfrac{x^3}{\sqrt{y^2+3}}\ge\sum\dfrac{x^3}{\sqrt{y^2+xy+yz+zx}}\ge\sum\dfrac{x^3}{\sqrt{\left(x+y\right)\left(z+y\right)}}\\ \overset{Cosi}{\ge}\sum\dfrac{2x^3}{x+2y+z}\ge2\sum\dfrac{\left(x^2\right)^2}{x^2+2xy+xz}\\ \overset{Svacxo}{\ge}2\dfrac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

\(\overset{Cosi}{\ge}\dfrac{2\left(x^2+y^2+z^2\right)^2}{4\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{2}\\ \overset{Cosi}{\ge}\dfrac{xy+yz+zx}{2}\ge\dfrac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

\(xy+yz+xz=xyz\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Áp dụng BĐT Cauchy Schwarz:

\(\dfrac{1}{4x+3y+z}\le\dfrac{1}{64}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

CMTT\(\Rightarrow\) \(M\le\dfrac{1}{64}\left(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\right)=\dfrac{1}{8}\)

Dấu''=" xảy ra\(\Leftrightarrow x=y=z=3\)

Tham khảo tại đây:

Câu hỏi của Hồ Minh Phi - Toán lớp 9 | Học trực tuyến

Lời giải:

Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)

Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)

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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)

\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)

\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)

\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)

Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)

Suy ra:

\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)

\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi $x=y=z=1$