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Ta có: \(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c.1+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{c\left(b+c\right)+a\left(b+c\right)}}=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{a}{a+c}.\frac{b}{b+c}}\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)\)( bđt Cosi)
Tương tự như trên: \(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right);\sqrt{\frac{ac}{b+ac}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{c}{b+c}\right)\)
=> \(P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{3}{2}\)
"=" Xảy ra khi và chỉ khi:
\(\frac{a}{a+c}=\frac{b}{b+c}\Leftrightarrow a\left(b+c\right)=b\left(a+c\right)\Leftrightarrow a=b\)
\(\frac{a}{a+b}=\frac{c}{b+c}\Leftrightarrow a=c\)
\(\frac{c}{a+c}=\frac{b}{a+b}\Leftrightarrow b=c\)
\(a+b+c=1\)
Từ các điều trên ta có đc: \(a=b=c=\frac{1}{3}\)
Vậy GTLN của P=3/2 khi và chỉ khi a=b=c=1/3
Ta có \(P=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(a+b\le2\sqrt{2}\) \(\Rightarrow\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)
Hay \(P=\frac{1}{a}+\frac{1}{b}\ge\sqrt{2}\)
Dấu "=" xảy ra <=> \(a=b=\sqrt{2}\)
Vậy \(P_{min}=\sqrt{2}\) tại \(a=b=\sqrt{2}\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
\(P=\frac{1}{\sqrt{a+3}}+\frac{1}{\sqrt{b+3}}\ge\frac{\left(1+1\right)^2}{\sqrt{a+3}+\sqrt{b+3}}=\frac{4}{\sqrt{a+3}+\sqrt{b+3}}\)
Ta có:
\(\left(\sqrt{a+3}.1+\sqrt{b+3}.1\right)^2\le\left(1^2+1^2\right)\left(a+3+b+3\right)\le16\)
\(\Rightarrow\sqrt{a+3}+\sqrt{b+3}\le\sqrt{16}=4\)
\(\Rightarrow P\ge\frac{4}{\sqrt{a+3}+\sqrt{b+3}}\ge\frac{4}{4}=1\).
Dấu \(=\)xảy ra khi \(a=b=1\).
\(P=\frac{1}{\sqrt{a+3}}+\frac{1}{\sqrt{b+3}}\Rightarrow P^2=\left(\frac{1}{a+3}+\frac{1}{b+3}\right)+\frac{2}{\sqrt{\left(a+3\right)\left(b+3\right)}}\)\(\ge\frac{4}{\left(a+b\right)+6}+\frac{2}{\frac{\left(a+3\right)+\left(b+3\right)}{2}}=\frac{8}{a+b+6}\ge\frac{8}{2+6}=1\)
\(\Rightarrow P\ge1\)
Đẳng thức xảy ra khi a = b = 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 phải chứng minh
\(\displaystyle \sum\)\(\frac{1+a}{b+c}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\)\(\displaystyle \sum\)\(\frac{2a+b+c}{b+c}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\)\(\displaystyle \sum\)\(\frac{2a}{b+c}+3\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\frac{a}{b}-\frac{a}{b+c}+\frac{b}{c}-\frac{b}{b+c}+\frac{c}{a}-\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{bc}{a\left(a+b\right)}+\frac{ab}{c\left(c+a\right)}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\left(ac\right)^2}{abc\left(b+c\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}+\frac{\left(ca\right)^2}{abc\left(c+a\right)}\ge\frac{3}{2}\)
Mặt khác: Theo BĐT AM-GM ta có:
\(\left(ab+bc+ca\right)^2\ge3\left(a^2bc+ab^2c+abc^2\right)=3abc\left(a+b+c\right)\)
Theo BĐT Cauchy-Schwwarz ta có:
\(\frac{\left(ac\right)^2}{abc\left(a+b\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}+\frac{\left(ca\right)^2}{abc\left(c+a\right)}\ge\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\frac{3}{2}\)
Bài toán được chứng minh xong. Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Từ bất đẳng thức luôn đúng \(\left(a-b\right)^2\ge0\)\(\Leftrightarrow a^2-2ab+b^2\ge0\)\(\Leftrightarrow a^2+b^2\ge2ab\)\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)(*)
Vì a, b là các số thực dương nên nhân cả 2 vế của (*) cho \(\frac{1}{ab\left(a+b\right)}\), ta có:
\(\frac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\frac{4}{ab\left(a+b\right)}\)\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\(\Leftrightarrow P\ge\frac{4}{a+b}\)
Lại có \(a+b\le2\sqrt{2}\)\(\Leftrightarrow\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)
Từ đó ta có \(P\ge\sqrt{2}\)
Dấu "=" xảy ra khi \(a=b=\sqrt{2}\)