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\(\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}=\frac{1}{a^2+a^2+b^2}+\frac{1}{b^2+b^2+c^2}+\frac{1}{c^2+c^2+a^2}\)
\(< =\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{9}\left(\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{1}{9}\left(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\)(bđt svacxo)
\(=\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)=\frac{1}{9}\cdot3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(=\frac{1}{9}\cdot3\cdot\frac{1}{3}=\frac{1}{9}\cdot1=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}< =\frac{1}{9}\)(đpcm)
dấu = xảy ra khi \(\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{1}{9}\Rightarrow a=b=c=3\)
ta có:\(\left(a+2b\right)^2=\left(1.a+\sqrt{2}.\sqrt{2}b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\)( bđt bunhiacopxki)
\(\left(a+2b\right)^2\le3.3c^2=9c^2\)→\(a+2b\le3c\)
lại có:\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
dấu = xảyra khi.... a+2b2=3c2(:v)
Câu 1: a)
b) Áp dụng Bđt Holder ta có:
\(\Rightarrow9\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{3}\ge\frac{\left(a+b+c\right)^3}{27}=\left(\frac{a+b+c}{3}\right)^3\)(đpcm)
Dấu = khi a=b=c
Câu 2:
Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)ta có:
\(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+b+1+1}=\frac{4}{3}\)(Đpcm)
Dấu = khi \(a=b=\frac{1}{2}\)
Câu 3:
Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\left(a+b+c=1\right)\)(Đpcm)
Dấu = khi \(a=b=c=\frac{1}{3}\)
Câu 4: nghĩ sau
Ta có \(\frac{a}{a^2+2b+3}=\frac{a}{a^2+1+2\left(b+1\right)}\le\frac{a}{2a+2\left(b+1\right)}=\frac{a}{2\left(a+b+1\right)}\)
Chứng minh tương tự \(\hept{\begin{cases}\frac{b}{b^2+2c+3}\le\frac{b}{2\left(b+c+1\right)}\\\frac{c}{c^2+2a+3}\le\frac{c}{2\left(a+c+1\right)}\end{cases}}\)
Cộng 3 vế của 3 bđt lại ta được
\(VT\le\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)
Để bài toán được chứng minh thì ta cần \(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)
\(\Leftrightarrow1-\frac{a}{a+b+1}+1-\frac{b}{b+c+1}+1-\frac{c}{c+a+1}\ge2\)
\(\Leftrightarrow A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge2\)
Ta có \(A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\)
\(=\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\frac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\frac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\)
Áp dụng bđt quen thuộc \(\frac{m^2}{x}+\frac{n^2}{y}+\frac{p^2}{z}\ge\frac{\left(m+n+p\right)^2}{x+y+z}\)(quen thuộc) ta được
\(A\ge\frac{\left(a+b+c+3\right)^2}{\left(b+1\right)\left(a+b+1\right)+\left(c+1\right)\left(b+c+1\right)+\left(a+1\right)\left(c+a+1\right)}\)
\(=\frac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)
\(=\frac{2\left(a+b+c+3\right)^2}{2\left(a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\right)}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+6}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9}\)
\(=\frac{2\left(a+b+c+3\right)^2}{\left(a+b+c+3\right)^2}=2\)(DDpcm)
Dấu "=" xảy ra tại a= b = c =1
bn có thể ghi cho mk cái bđt đấy đc ko
#mã mã#
Ta có :(a+b-c)2 \(\ge\) 0
<=>a2+b2+c2 \(\ge\) 2(bc-ab+ac)
<=>\(\frac{5}{3}\ge\) 2(bc-ab+ac)
<=>bc+ac-ab \(\le\frac{5}{6}< 1\)
<=>\(\frac{bc+ac-ab}{abc}< \frac{1}{abc}\) (vì a,b,c>0 nên chia cả 2 vế cho abc)
<=>\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< 1\) (đpcm)
Sửa lại đề là \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{a^2+10}+\frac{4}{b^2+10}+\frac{4}{c^2+10}\) nhé
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ( a,b > 0 ) , ta có :
\(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\); \(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{b+2c+a}\);
\(\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{4}{c+2a+b}\) ( I )
Lại có : \(\frac{1}{2a+b+c}\ge\frac{2}{2a^2+b^2+c^2+4}=\frac{2}{a^2+10}\)
tương tự \(\frac{1}{2b+c+a}\ge\frac{2}{b^2+10}\); \(\frac{1}{2c+a+b}\ge\frac{2}{c^2+10}\)( II )
Từ (I) và (II) => \(2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{8}{a^2+10}+\frac{8}{b^2+10}+\frac{8}{c^2+10}\)
=> \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{a^2+10}+\frac{4}{b^2+10}+\frac{4}{c^2+10}\)( đpcm )
Dấu "=" xảy ra <=> a = b = c = \(\frac{\sqrt{21}}{3}\)