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Các câu hỏi tương tự
Cho x,y,z>0 thoã mãn: x3+y3+z3=1
Chứng minh rằng: \(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)
Cho x, y, z 0 thoả mãn x+y+z1. 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}}gesqrt{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}}gesqrt{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}}gesqrt{406}
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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 thoả mãn x+y+z2. Tìm GTNN của các biểu thức:
a) Asqrt{x^2+dfrac{1}{x^2}}+sqrt{y^2+dfrac{1}{y^2}}+sqrt{z^2+dfrac{1}{z^2}}
b) Bsqrt{x^2+dfrac{1}{y^2}+dfrac{1}{z^2}}+sqrt{y^2+dfrac{1}{z^2}+dfrac{1}{x^2}}+sqrt{z^2+dfrac{1}{x^2}+dfrac{1}{y^2}}
c) Csqrt{2x^2+dfrac{3}{y^2}+dfrac{4}{z}}+sqrt{2y^2+dfrac{3}{z^2}+dfrac{4}{x^2}}+sqrt{2z^2+dfrac{3}{x^2}+dfrac{4}{y^2}}
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Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
Cho các số dương x,y,z thỏa mãn \(x^3+y^3+z^3=1\). CMR :
\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)
Cho x\(\ge\)3; y\(\ge2\); z\(\ge\)1. Chứng minh rằng:
\(\dfrac{xy\sqrt{x-1}+zx\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)
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
28 tháng 6 2021 lúc 20:17
Cho các số thực dương thoả mãn: \(x\sqrt{1-y^2}+y\sqrt{1-z^2}+z\sqrt{1-x^2}=\dfrac{3}{2}\)
Cmr: \(x^2+y^2+z^2=\dfrac{3}{2}\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
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}ge3sqrt[3]{dfrac{x^3}{y+z}.dfrac{y+z}{36}.dfrac{1}{162}}+3sqrt[3]{dfrac{y^3}{x+z}.dfrac{x+z}{36}.dfrac{1}{162}}+3sqrt[3]{dfrac{z^3}{x+y}.dfrac{x+y}{36}.dfrac{1}{162}}3sqrt[3]{dfrac{x^3}{36.162}}+3sqrt[3]{dfrac{y^3}{36.162}}+3sqrt[3]{dfrac{z^3}{...
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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=\(\dfrac{1}{9}\)