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+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) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
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=\(\dfrac{1}{9}\)
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
do x,y,z≥0 nên x2≥0 , y+z≥0
áp dụng bất đẳng thức cosi cho 2 số dương \(\dfrac{x^2}{y+z}\) và y+z/4
x^2/y+z +(y+z)/4≥2\(\sqrt{\dfrac{x^2}{y+z}.\dfrac{\left(y+z\right)}{4}}\) =x (1)
y^2/x+z+(x+z)/4≥2\(\sqrt{\dfrac{y^2}{x+z}.\dfrac{x+z}{4}}\) =y (2)
z^2/y+x+(y+x)/4≥2\(\sqrt{\dfrac{z^2}{y+x}.\dfrac{y+x}{4}}\) =z (3)
từ (1)(2)(3)
➜\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) +(a+b+c)/2 ≥x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥ (x+y+z)/2
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥1 (vì x+y+z=2)
vậy giá trị nhỏ nhất của \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) =1
cho x,y,z>0 x+y+z<=1
tìm GTNN:P= \(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
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 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\(\ge\)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 x,y,z>0 thoã mãn: x3+y3+z3=1
Chứng minh rằng: \(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)
