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Ta sẽ chứng minh :
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) với x, y > 0
Thật vậy : \(x+y+z\ge3\sqrt[3]{xyz}\)( bđt Cô - si )
Và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\) ( bđt Cô - si )
\(\Rightarrow x+y+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( Dấu " = " \(\Leftrightarrow x=y=z\) )
Ta có :
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
( Dấu " = " xay ra khi a=b)
Tương tự ta cũng có :
\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\) ( Dấu " = " xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\) ( Dấu " = " xay ra khi c = a )
\(VT=\sum_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu " = " xay ra khi \(a=b=c=\frac{2}{3}\)
Chúc bạn học tốt !!
\(\frac{1}{\sqrt{4a^2+2ab+b^2+a^2+b^2}}\le\frac{1}{\sqrt{4a^2+2ab+b^2+2ab}}=\frac{1}{\sqrt{\left(2a+b\right)^2}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow VT\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự ta có: \(\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}\le\dfrac{1}{9}\left(\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế với vế:
\(\dfrac{1}{\sqrt{5a^2+2ab+b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+c^2}}+\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\le\dfrac{2}{3}\)
Dấu "=" khi \(a=b=c=\dfrac{3}{2}\)
với mọi x,y,z >0 ta có: \(x+y+z\ge3\sqrt[3]{xyz};\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
\(\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
đẳng thức xảy ra khi x=y=z
ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
đẳng thức xảy ra khi a=b
tương tự: \(\frac{1}{\sqrt{5b^2+2ab+2b^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
đẳng thức xảy ra khi b=c
\(\frac{1}{\sqrt{5c^2+2bc+2c^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
đẳng thức xảy ra khi c=a
Vậy \(\frac{1}{\sqrt{5a^2+2ca+2a^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
đẳng thức xảy ra khi a=b=c=\(\frac{3}{2}\)
Ta có: \(5a^2+2ab+2b^2=4a^2+2ab+b^2+\left(a^2+b^2\right)\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Lại có: \(\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
Tương tự cộng lại ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo BĐT Bunhiacopxki ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{3}\)
\(\Rightarrow VT\le\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)
Dấu = xảy ra khi \(a=b=c=\sqrt{3}\)
Bài 1 :
a ) + Ta có \(\Delta ABE\) vuông tại E và \(\Delta ACF\) vuông tại F ( vì BE và CF là hai đường cao của\(\Delta\)ABC )
\(\Rightarrow cosBAC=\frac{AE}{AB}=\frac{AF}{AC}\Rightarrow AE.AC=AF.AB\)
+ ) Ta có : \(\Delta ADC\) vuông tại D có DK là đường cao
\(\Rightarrow AD^2=AK.AC\)
Lại có : \(\Delta ADB\) vuông tại D có DI là đường cao
\(\Rightarrow AD^2=AI.AB\)
Suy ra : \(AI.AB=AK.AC\)
b ) Ta có : \(\Delta ADB\) vuông tại D \(\Rightarrow\sin ABC=\frac{AD}{AB}\)
Lại co : \(\Delta CBE\) vuông tại E và \(\Delta AHE\) vuông tại E
Mà \(\widehat{AHE}=\widehat{C}\) ( cùng bù ^DHE ) \(\Rightarrow\sin ABC=\frac{BE}{BC}=\frac{AE}{AH}\)
\(\Rightarrow\frac{\cos BAC}{\sin ABC.\sin ACB}=\frac{AE}{AB}:\left(\frac{AD}{AB}.\frac{AE}{AH}\right)=\frac{AE}{AB}.\frac{AB.AH}{AD.AE}=\frac{AH}{AD}\)
Vậy \(AD.cosBAC=AH.\sin ABC.\sin ACB\left(đpcm\right)\)
\(\frac{1}{\sqrt{5a^2+2ab+2b^2}}=\frac{1}{\sqrt{4\left(a+\frac{b}{2}\right)^2+\left(a-b\right)^2}}\le\frac{1}{\sqrt{4\left(a+\frac{b}{2}\right)^2}}=\frac{1}{2\left(a+\frac{b}{2}\right)}=\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
CM tương tự => \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{9}\left(\frac{2}{b}+\frac{1}{c}\right)\)
\(\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{2}{c}+\frac{1}{a}\right)\)
Cộng vế với vế => \(\frac{1}{\sqrt{5a^2+2ab+2b^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2c^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu "=" xảy ra <=> \(a=b=c=\frac{3}{2}\)
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
Ta sẽ chứng minh: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)với x,y > 0.
Thật vậy: \(x+y+z\ge3\sqrt[3]{xyz}\)(bđt Cô -si)
và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\)(bđt Cô -si)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)(Dấu "="\(\Leftrightarrow x=y=z\))
Ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
(Dấu "=" xảy ra khi a = b)
Tương tự ta có:\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)(Dấu "=" xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)(Dấu "=" xảy ra khi c=a)
\(VT=\text{Σ}_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
(Dấu "=" xảy ra khi \(a=b=c=\frac{3}{2}\))
Ô, thanh you, bạn 2k7 sao mà giỏi thế