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\(\frac{1}{1-ab}=1+\frac{ab}{1-ab}\le1+\frac{ab}{1-\frac{a^2+b^2}{2}}=1+\frac{2ab}{2-a^2-b^2}=1+\frac{2ab}{2c^2+a^2+b^2}\)
\(=1+\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le1+\frac{ab}{\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}}=1+\sqrt{\frac{a^2b^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}}\)
\(\le1+\frac{1}{2}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)\)
Áp dụng BĐT Cô si, ta có:
\(\begin{aligned} \frac{1}{1-ab}&=1+\frac{ab}{1-ab} \le 1+\frac{ab}{1-\frac{a^2+b^2}{2}}=1+\frac{2ab}{a^2+b^2+2c^2} \\ &=1+\frac{2ab}{(a^2+c^2)+(b^2+c^2)}\le 1+\frac{ab}{\sqrt{(a^2+c^2)(b^2+c^2)}}\\& \le 1+\frac{1}{2}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right). \text{ }(1)\end{aligned}\)
Tương tự \(\frac{1}{1-bc}\le1+\frac{1}{2}\left(\frac{b^2}{b^2+a^2}+\frac{c^2}{a^2+c^2}\right)\left(2\right)\)
\(\frac{1}{1-ca}\le1+\frac{1}{2}\left(\frac{c^2}{c^2+b^2}+\frac{a^2}{a^2+b^2}\right)\left(3\right)\)
\(\Rightarrow VT\le3+\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{9}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
=> \(-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6\)
=> \(-\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6.\frac{3}{2}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
=> \(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\)
=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(1)
Dễ thấy \(\frac{a}{b}+\frac{b}{a}\ge2\)(với a,b > 0)
=> (1) đúng
=> BĐTđược chứng minh
b)Đặt \(A=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(a,b,c>0\right)\).
\(A=4\left(a+b+c\right)-3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\).
\(A=\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\).
Vì \(a>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(4a+\frac{1}{a}\ge2\sqrt{4.a.\frac{1}{a}}=4\left(1\right)\).
Dấu bằng xảy ra \(\Leftrightarrow4a=\frac{1}{a}\Leftrightarrow a=\frac{1}{2}\).
Chứng minh tương tự, ta được:
\(4b+\frac{1}{b}\ge4\left(b>0\right)\left(2\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=\frac{1}{2}\).
Chứng minh tương tự, ta được:
\(4c+\frac{1}{c}\ge4\left(c>0\right)\left(3\right)\).
Dấu bằng xảy ra \(\Leftrightarrow c=\frac{1}{2}\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)\ge4+4+4=12\).
\(\Leftrightarrow\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\ge\)\(12-3\left(a+b+c\right)\).
\(\Leftrightarrow A\ge12-3\left(a+b+c\right)\left(4\right)\).
Mặt khác, ta có: \(a+b+c\le\frac{3}{2}\).
\(\Leftrightarrow3\left(a+b+c\right)\le\frac{9}{2}\).
\(\Rightarrow-3\left(a+b+c\right)\ge-\frac{9}{2}\).
\(\Leftrightarrow12-3\left(a+b+c\right)\ge\frac{15}{2}\left(5\right)\).
Dấu bằng xảy ra \(\Leftrightarrow a+b+c=\frac{3}{2}\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(A\ge\frac{15}{2}\).
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\).
Vậy với \(a,b,c>0\)và \(a+b+c\le\frac{3}{2}\)thì \(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{15}{2}\).
Ta có : \(\frac{1}{1-ab}=1+\frac{ab}{1-ab}\le1+\frac{ab}{1-\frac{a^2+b^2}{2}}=1+\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\)
\(\le1+\frac{a.b}{\sqrt{a^2+c^2}.\sqrt{b^2+c^2}}\le1+\frac{1}{2}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)\)
Tương tự , ta chứng minh được \(\frac{1}{1-bc}\le1+\frac{1}{2}\left(\frac{b^2}{b^2+a^2}+\frac{c^2}{c^2+a^2}\right)\)
\(\frac{1}{1-ac}\le1+\frac{1}{2}\left(\frac{a^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}\right)\)
Cộng theo vế : \(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\le3+\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{9}{2}\)