Cho a, b > 0. CM: \(\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge\sqrt{ab}+\frac{a+b}{2}\)
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a/
\(=\frac{a+b}{b^2}.\frac{\left|a\right|.b^2}{\left|a+b\right|}=\frac{\left(a+b\right).b^2.\left|a\right|}{b^2\left(a+b\right)}=\left|a\right|\)
b/
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{2\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
BĐT<=>
\(\left(\frac{2ab}{a+b}-\frac{a+b}{2}\right)+\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)\ge0\)
<=> \(-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\frac{a^2+b^2}{2}-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}\ge0\)
<=> \(\frac{\left(a-b\right)^2}{2(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab})}-\frac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)
<=> \(a+b\ge\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\)
<=> \(\frac{a^2+b^2}{2}+ab\ge2\sqrt{\frac{a^2+b^2}{2}.ab}\)luôn đúng
=> ĐPCM
Dấu bằng xảy ra khi a=b
a/ \(\frac{b}{b}.\sqrt{\frac{a^2+b^2}{2}}+\frac{c}{c}.\sqrt{\frac{b^2+c^2}{2}}+\frac{a}{a}.\sqrt{\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{b}.\left(\frac{3b^2+a^2}{4}\right)+\frac{1}{c}.\left(\frac{3c^2+b^2}{4}\right)+\frac{1}{a}.\left(\frac{3a^2+c^2}{4}\right)\)
\(=\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\)
Ta cần chứng minh
\(\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\left(a+b+c\right)\)
Mà: \(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Vậy có ĐPCM.
Câu b làm y chang.
Bất đẳng thức cần chứng minh tương đương với:
\(\frac{a+b}{\sqrt{ab}}\)+\(\frac{4\sqrt{ab}}{a+b}\)-\(\frac{3ab}{a+b}\)\(\ge\)\(\frac{5}{2}\)(*)
Nhưng mà theo bất đẳng thức AM-GM thì (*) tương đương với
2\(\sqrt{\frac{a+b}{\sqrt{ab}}.\frac{4\sqrt{ab}}{a+b}}\)-\(\frac{3\sqrt{ab}}{2\sqrt{ab}}\)\(\ge\)\(\frac{5}{2}\)
và tương đương với 4-\(\frac{3}{2}\)\(\ge\)\(\frac{5}{2}\)hiển nhiên đúng nên (*) đúng hay ta có đpcm
Vậy \(\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)\(\ge\)\(\frac{5}{2}\)
dấu đẳng thức xảy ra khi a=b
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 \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)