Chủ đề này có 2 trả lời

[L Kira]

Cho $x,y,z>0$ : $x+y+z=3$.Chứng minh: 

$$\frac{4x+5}{x^{3}+xy^{2}+3xyz}+\frac{4y+5}{y^{3}+yz^{2}+3xyz}+\frac{4z+5}{z^{3}+zx^{2}+3xyz}\geq\frac{162}{x^{2}+y^{2}+z^{2}+27}$$

[Hoang Tung 126]

Cho $x,y,z>0$ : $x+y+z=3$.Chứng minh: 

$$\frac{4x+5}{x^{3}+xy^{2}+3xyz}+\frac{4y+5}{y^{3}+yz^{2}+3xyz}+\frac{4z+5}{z^{3}+zx^{2}+3xyz}\geq\frac{162}{x^{2}+y^{2}+z^{2}+27}$$

Áp dụng bđt Cosi ta có :$4x+5=x+x+x+x+1+1+1+1+1\geq 9\sqrt[9]{x^4}= > 4x+5\geq 9\sqrt[9]{x^4}= > \sum \frac{4x+5}{x^3+xy^2+3xyz}=\sum \frac{4x+5}{x(x^2+y^2+3yz)}\geq \sum \frac{9\sqrt[9]{x^4}}{x(x^2+y^2+3yz)}=9\sum \frac{1}{\sqrt[9]{x^5}(x^2+y^2+3yz)}\geq 9.\frac{3}{\sqrt[3]{\sqrt[9]{(xyz)^5}(x^2+y^2+3yz)(y^2+z^2+3xz)(x^2+z^2+3xy)}}$     (3)

 

 Mà $3=x+y+z\geq 3\sqrt[3]{xyz}= > xyz\leq 1= > \sqrt[9]{(xyz)^5}\leq 1$    (1)

 

  $\sqrt[3]{(x^2+y^2+3yz)(y^2+z^2+3xz)(z^2+x^2+3xy)}\leq \frac{x^2+y^2+3yz+y^2+z^2+3xz+z^2+x^2+3xy}{3}=\frac{2(x^2+y^2+z^2)+3(xy+yz+xz)}{3}$      (2)

 

 Từ (1),(2) $= > \sqrt[3]{\sqrt[9]{(xyz)^5}(x^2+y^2+3yz)(y^2+z^2+3xz)(z^2+x^2+3xy)}\leq \frac{2(x^2+y^2+z^2)+3(xy+yz+xz)}{3}=\frac{4(x^2+y^2+z^2)+6(xy+yz+xz)}{6}=\frac{x^2+y^2+z^2+3(x+y+z)^2}{6}=\frac{x^2+y^2+z^2+3.3^2}{6}=\frac{x^2+y^2+z^2+27}{6}$       (4)

 

 Từ (3),(4) $= > \sum \frac{4x+5}{x^3+xy^2+3xyz}\geq 9.\frac{3}{\frac{x^2+y^2+z^2+27}{6}}=\frac{162}{x^2+y^2+z^2+27}$

 

 Do đó ta có ĐPCM .Dấu = xảy ra khi $x=y=z=1$

[NguyenDangHuyYTNA]

$Taco:\sum \frac{4x+5}{x(x^2+y^2+3yz)}=\sum \frac{9(4+\frac{5}{x})}{9(x^2+y^2+3yz)}\geq \sum \frac{(4+\frac{5}{\sqrt{x}})^2}{9(x^2+y^2+3yz)}\geq \frac{(12+\frac{45}{3})^2}{9(2\sum x^2+3\sum xy)}=\frac{162}{x^2+y^2+z^2+27(dpcm)}$