Cho x,y,z>0 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\). Tìm MinP \(=x+\dfrac{y^2}{2}+\dfrac{z^3}{3}\)
cho các số thực x,y,z thoả mãn x+y+z≥6.
Tìm minP=\(\dfrac{x^2}{yz+\sqrt{1+x^3}}+\dfrac{y^2}{xz+\sqrt{1+y^3}}+\dfrac{z^2}{xy+\sqrt{1+z^3}}\)
Cho mng tham khảo ạ
Cho x,y,z>0 và \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1\).Tìm MinP = \(\Sigma\dfrac{x^3}{y\left(x+z\right)}\)
Cho x,y,z>0 . Tìm MinP = \(\Sigma\dfrac{x^2}{y^2+yz+z^2}\)
Cho x,y,z> 0. Tìm MinP = \(\Sigma\dfrac{x}{\sqrt{x^2+8yz}}\)
Cho x,y,z>0 và \(xy\sqrt{xy}+yz\sqrt{yz}+xz\sqrt{xz}=1\)
Tìm MinP= \(\Sigma\dfrac{x^6}{x^3+y^3}\)
Cho x, y, z > 0 thoả mãn x+y+z=1. Chứng minh rằng:
a) \(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\sqrt{82}\)
b) \(\sqrt{x^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}\ge\sqrt{163}\)
c)\(\sqrt{x^2+\dfrac{2}{y^2}+\dfrac{3}{z^2}}+\sqrt{y^2+\dfrac{2}{z^2}+\dfrac{3}{x^2}}+\sqrt{z^2+\dfrac{2}{z^2}+\dfrac{3}{y^2}}\ge\sqrt{406}\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Cho x,y,z>0 và \(xy+yz+xz\ge3\)
Tìm MinP = \(\Sigma\dfrac{x^3}{\sqrt{y^2+3}}\)
Đề bài: ax,y,z >0 và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\). Tìm Min P= \(\dfrac{x^3}{y+z}+\dfrac{y^3}{z+x}+\dfrac{z^3}{x+y}\).
ĐÁP ÁN:
Ta có: \(\dfrac{x^3}{y+z}+\dfrac{y+z}{36}+\dfrac{1}{162}+\dfrac{y^3}{x+z}+\dfrac{x+z}{36}+\dfrac{1}{162}+\dfrac{z^3}{x+y}+\dfrac{x+y}{36}+\dfrac{1}{162}\ge3\sqrt[3]{\dfrac{x^3}{y+z}.\dfrac{y+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{y^3}{x+z}.\dfrac{x+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{z^3}{x+y}.\dfrac{x+y}{36}.\dfrac{1}{162}}=3\sqrt[3]{\dfrac{x^3}{36.162}}+3\sqrt[3]{\dfrac{y^3}{36.162}}+3\sqrt[3]{\dfrac{z^3}{36.162}}=\dfrac{x+y+z}{6}.\)
=> P+\(\dfrac{x+y+z}{18}+\dfrac{1}{54}\)≥\(\dfrac{x+y+z}{6}\) <=> P≥\(\dfrac{x+y+z}{6}-\dfrac{x+y+z}{18}-\dfrac{1}{54}\)=\(\dfrac{x+y+z}{9}-\dfrac{1}{54}\)
Ta c/m đc: 3(x+y+z)≥(\(\sqrt{x}+\sqrt{y}+\sqrt{z}\))2 <=> 2(x+y+z) ≥2\(\left(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\right)\)<=> x+y+z≥\(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\)(luôn đúng)
➩x+y+z ≥ \(\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^3}{3}=\dfrac{1}{3}\) => P≥\(\dfrac{1}{54}\). Dấu ''='' xảy ra <=> x=y=z=\(\dfrac{1}{9}\)
