Cho a,b>0.tìm minP= \(\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
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12 tháng 5 2019
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left[\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right].\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\left[\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(a-b\right)\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{a-b}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
LM
4 tháng 1 2020
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}\\ =\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right):\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\\ =\left(\frac{\sqrt{a^2}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{b^2}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right):\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}(\sqrt{a}-\sqrt{b})}\\ =\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}\)
9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
29 tháng 6 2019
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{\sqrt{a^3}-3a\sqrt{b}+3\sqrt{a}.b-\sqrt{b^3}+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{3\sqrt{a^3}-3a\sqrt{b}+3b\sqrt{a}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
KT
23 tháng 7 2018
a) ĐK: a > 0; b > 0
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}-b\)
\(=\frac{\sqrt{a}+\sqrt{b}+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}-b\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)-b\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}-b\)
\(=2\sqrt{b}-b\)
b) \(A=1\)\(\Rightarrow\)\(2\sqrt{b}-b=1\)
\(\Leftrightarrow\)\(b-2\sqrt{b}+1=0\)
\(\Leftrightarrow\) \(\left(\sqrt{b}-1\right)^2=0\)
\(\Leftrightarrow\)\(\sqrt{b}-1=0\)
\(\Leftrightarrow\)\(\sqrt{b}=1\)
\(\Leftrightarrow\)\(b=1\) (t/m ĐKXĐ)
Vậy b=1
20 tháng 5 2019
Ta có \(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{\left(c+b\right)\left(c+a\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{a+b}\right)\)
Khi đó \(P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\frac{1}{2}\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{c}{b+c}\right)=\frac{3}{2}\)
\(MaxP=\frac{3}{2}\)khi a=b=c=1/3
15 tháng 11 2017
ta có : \(P=\frac{\sqrt{bc}}{a+2\sqrt{bc}}+\frac{\sqrt{ac}}{b+2\sqrt{ac}}+\frac{\sqrt{ab}}{c+2\sqrt{ab}}\le\frac{\frac{1}{2}\left(b+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+b\right)}{a+b+c}\)
\(\Rightarrow P\le\frac{a+b+c}{a+b+c}=1\)
=> GTLN của P là 1 khi a=b=c
T
22 tháng 5 2019
Ta có:
\(\frac{ab}{\sqrt{2017c+ab}}=\frac{ab}{\sqrt{\left(a+b+c\right)c+ab}}\)
\(=\frac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
Áp dụng BĐT AM-GM (cô si): \(ab.\frac{1}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{ab}{2}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{ab}{2\left(a+c\right)}+\frac{ab}{2\left(b+c\right)}\)
Tương tự với hai BĐT còn lại và cộng theo vế,ta được:
\(A\le\frac{ab}{2\left(a+c\right)}+\frac{ab}{2\left(b+c\right)}+\frac{bc}{2\left(a+b\right)}+\frac{bc}{2\left(a+c\right)}+\frac{ca}{2\left(b+c\right)}+\frac{ca}{2\left(a+b\right)}\)
Thu gọn lại bằng cách cộng những phân thức cùng mẫu và rút gọn phân thức,ta được:
\(A\le\frac{a+b+c}{2}=\frac{2017}{2}\).
Dấu "=" xảy ra khi \(a=b=c=\frac{2017}{3}\)
Vậy...
TP
0
Ta có:
\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
\(P=\left(\frac{\sqrt{ab}}{a+b}+\frac{a+b}{4\sqrt{ab}}\right)+\frac{3\left(a+b\right)}{4\sqrt{ab}}\)
\(\ge2\sqrt{\frac{\sqrt{ab}}{a+b}\cdot\frac{a+b}{4\sqrt{ab}}}+\frac{3\cdot2\sqrt{ab}}{4\sqrt{ab}}\) (BĐT Cauchy)
\(=2\cdot\frac{1}{2}+\frac{3}{2}=\frac{5}{2}\)
Dấu "=" xảy ra khi: a = b
Vậy \(Min_P=\frac{5}{2}\Leftrightarrow a=b\)
\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3}{4}\frac{a+b}{\sqrt{ab}}+\frac{1}{4}\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
Ta có:
\(a+b\ge2\sqrt{ab}\Leftrightarrow\frac{a+b}{\sqrt{ab}}\ge2\).
\(\frac{1}{4}\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\ge2\sqrt{\frac{1}{4}\frac{\left(a+b\right)\sqrt{ab}}{\sqrt{ab}\left(a+b\right)}}=1\).
Suy ra \(P\ge\frac{3}{4}.2+1=\frac{5}{2}\).
Dấu \(=\)xảy ra khi \(a=b\).