theo bđt cauchy schwarz ta có
\(12\sqrt{6-x}\le\dfrac{12^2+6-x}{2}\)
\(5\sqrt{x-5}\le\dfrac{25+x-5}{2}\)
\(\Rightarrow12\sqrt{6-x}+5\sqrt{x-5}\le\dfrac{144+6-x+25+x-5}{2}=85\)
vậy Amax=85
theo bđt cauchy schwarz ta có
\(12\sqrt{6-x}\le\dfrac{12^2+6-x}{2}\)
\(5\sqrt{x-5}\le\dfrac{25+x-5}{2}\)
\(\Rightarrow12\sqrt{6-x}+5\sqrt{x-5}\le\dfrac{144+6-x+25+x-5}{2}=85\)
vậy Amax=85
tìm max của A=-x+5\(\sqrt{x}\)-\(\frac{9}{\sqrt{x}}\)+2019
4) \(\dfrac{x-\sqrt{x}}{1-\sqrt{2\left(x^2-x+1\right)}}\ge1\)
5)\(x^2+x+1>3\sqrt{x}\left(x+1\right)\)
6)\(\dfrac{1}{1-x^2}>\dfrac{3x}{\sqrt{1-x^2}}-1\)
nữa ạ
Tìm GTLN của các biểu thức sau:
\(A=2x\left(6-x\right),0\le x\le6\)
\(B=x\sqrt{9-x},0\le x\le9\)
\(C=\left(6-x\right)\sqrt{x},0\le x\le6\)
với mọi x,y,z >0 CMR: \(\dfrac{1+\sqrt{x}}{y+z}+\dfrac{1+\sqrt{y}}{z+x}+\dfrac{1+\sqrt{z}}{x+y}\ge\dfrac{9+3\sqrt{3}}{2}\)
Cho x,y,z và xyz \(\ge\) 1. CMR: \(\dfrac{x}{\sqrt{x+\sqrt{yz}}}+\dfrac{y}{\sqrt{y+\sqrt{xz}}}+\dfrac{z}{\sqrt{z+\sqrt{xy}}}\ge\dfrac{3}{\sqrt{2}}\)
với mọi x, y, z dương thỏa mãn x+y+z =1: CMR: \(\dfrac{1+\sqrt{x}}{y+z}+\dfrac{1+\sqrt{y}}{z+x}+\dfrac{1+\sqrt{z}}{x+y}\ge\dfrac{9+3\sqrt{3}}{2}\)
Bài 1: Cho x,y, z > 0 thỏa mãn xyz = 1.
Chứng minh rằng:
\(\dfrac{\sqrt{1+x^3+y}^3}{xy}\)+ \(\dfrac{\sqrt{1+x^3+z^3}}{xz}\)+ \(\dfrac{\sqrt{1+y^3+z^3}}{yz}\) ≥ \(3\sqrt{3}\)
Bài 2: Choa, b, c,d > 0 thỏa mãn abcd = 1. CMR:
1) \(\dfrac{a^3}{c^6}\)+ \(\dfrac{c^3}{a^6}\)+ \(\dfrac{b^3}{d^6}\)+ \(\dfrac{d^3}{b^6}\) ≥ \(\dfrac{a^2}{c}\)+ \(\dfrac{c^2}{a}+\dfrac{b^2}{d}+\dfrac{d^2}{b}\)
2) \(\dfrac{a^5b^4}{c^{13}}\) + \(\dfrac{b^5c^4}{d^{13}}\) + \(\dfrac{c^5d^4}{a^{13}}\)+ \(\dfrac{d^5a^4}{b^{13}}\) ≥ \(\dfrac{ab^2}{c^3}+\dfrac{bc^2}{d^3}+\dfrac{cd^2}{a^3}\)+ \(\dfrac{da^2}{b^3}\)
Bài 3: Cho a, b,c ,d > 0. CMR:
\(\dfrac{a^2}{b^5}+\dfrac{b^2}{c^5}+\dfrac{c^2}{d^5}+\dfrac{d^2}{a^5}\) ≥ \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{1}{d^3}\)
Bài 4: tìm giá trị nhỏ nhất của biểu thức:
A= x + y biết x, y > 0 thỏa mãn \(\dfrac{2}{x}+\dfrac{3}{y}\) = 1
B= \(\dfrac{ab}{a^2+b^2}\) + \(\dfrac{a^2+b^2}{ab}\) với a, b > 0
Bài 5: Với x > 0, chứng minh rằng:
( x+2 )2 + \(\dfrac{2}{x+2}\) ≥ 3
Giúp mk với, mai mk phải kiểm tra rồi!!
Cho x,y,z>0 C/M
\(\sqrt{\dfrac{x^3}{y^3}}+\sqrt{\dfrac{y^3}{z^3}}+\sqrt{\dfrac{z^3}{x^3}}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\)
1) \(x+\sqrt{1-x^2}< x\sqrt{1-x^2}\)
2)\(\dfrac{1}{\sqrt{2x^2+3x-3}}>\dfrac{1}{2x-1}\)
3)\(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}< 2x+\dfrac{1}{2x}+4\)
giúp mình ạ