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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 \(a=b=c=\dfrac{3}{2}\)
Tìm trước khi hỏi :
Đề vòng 1 chuyên sư phạm 2016-2017 - Tài liệu - Đề thi - Diễn đàn Toán học
Witch Rose
Vì a,b,ca,b,c không âm và a+b+c=1a+b+c=1 nên 2≤t=√5c+4≤32≤t=5c+4≤3
Ta có:a,b≥0⇒25ab+20(a+b)+16≥20(a+b)+16a,b≥0⇒25ab+20(a+b)+16≥20(a+b)+16
⇔(5a+4)(5b+4)≥4(5a+5b+4)⇔(5a+4)(5b+4)≥4(5a+5b+4)
⇔(√5a+4+√5b+4)2≥(2+√5a+5b+4)2⇔(5a+4+5b+4)2≥(2+5a+5b+4)2
⇔√5a+4+√5b+4≥2+√9−5c=2+√13
\(S\ge3\frac{1}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right)}}.\)
\(S\ge\frac{3.4}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right).16.16.16}}\)
\(S\ge\frac{12}{\frac{a+2b+5c+b+2c+5a+c+2a+5b+16+16+16}{6}}\)
\(S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)
\(a;b;c\ge0;a+b+c=1\Rightarrow a;b;c\le1\)
\(\Rightarrow\left\{{}\begin{matrix}a^2\le a\\b^2\le b\\c^2\le c\end{matrix}\right.\)
\(\sqrt{5a+4}+\sqrt{5b+4}+\sqrt{5c+4}\)
\(=\sqrt{a+4a+4}+\sqrt{b+4b+4}+\sqrt{c+4c+4}\)
\(\ge\sqrt{a^2+4a+4}+\sqrt{b^2+4b+4}+\sqrt{c^2+4c+4}=\sqrt{\left(a+2\right)^2}+\sqrt{\left(b+2\right)^2}+\sqrt{\left(c+2\right)^2}\)
\(=a+b+c+2+2+2=7\)
\("="\Leftrightarrow a;b;c\) là hoán vị của (0;0;1)
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 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 \(a=b=c=\frac{3}{2}\))
Đặt VT= \(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\)
Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki ta có:
\(VT^2=\left(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\right)^2\le\)
\(\le\left(1^2+1^2+1^2\right)\left(a+b+c+ab+bc+ca\right)\)
Lại có \(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2\)( tự cm nhé)
Từ đó \(VT^2\le3.\left(1+\dfrac{1}{3}\right)=4\) (do a+b+c=1)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Áp dụng bđt bunhiacopxki có:
\(\left(\sqrt{5a+1}+\sqrt{5b+1}+\sqrt{5c+1}\right)^2\le\left(5a+1+5b+1+5c+1\right)\left(1^2+1^2+1^2\right)=3\cdot\left[5\left(a+b+c\right)+3\right]=3\cdot8=24\)
\(\Leftrightarrow\sqrt{5a+1}+\sqrt{5b+1}+\sqrt{5c+1}\le\sqrt{24}=2\sqrt{6}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{3}\)