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Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{2c}\right)=\left(x;y;z\right)\)
BĐT trở thành: \(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}\ge\frac{8}{\sqrt{x^2+y^2+\frac{z^2}{2}}}\)
Ta có: \(VT=\frac{1}{x}+\frac{1}{y}+\frac{2^2}{z}\ge\frac{\left(1+1+2\right)^2}{x+y+z}=\frac{16}{x+y+z}\) (1)
\(\left(1.x+1.y+\sqrt{2}.\frac{z}{\sqrt{2}}\right)^2\le\left(1+1+2\right)\left(x^2+y^2+\frac{z^2}{2}\right)\)
\(\Rightarrow x+y+z\le2\sqrt{x^2+y^2+\frac{z^2}{2}}\)
\(\Rightarrow VP=\frac{8}{\sqrt{x^2+y^2+\frac{z^2}{2}}}\le\frac{16}{x+y+z}\)(2)
Từ (1); (2) suy ra đpcm
Dấu "=" xảy ra khi \(x=y=\frac{z}{2}\) hay \(a=b=\frac{c}{2}\)
a) Bình phương 2 vế được: \(\frac{4ab}{a+b+2\sqrt{ab}}\le\sqrt{ab}\)
<=> \(4ab\le\sqrt{ab}\left(a+b\right)+2ab\)
<=>\(\sqrt{ab}\left(a+b\right)\ge2ab\)
<=>\(a+b\ge2\sqrt{ab}\)
<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
Vậy \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\forall a,b>0\)
Bài 1:
a) Áp dụng BĐT Cô-si:
\(VT=a-1+\frac{1}{a-1}+1\ge2\sqrt{\frac{a-1}{a-1}}+1=2+1=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=2\).
b) BĐT \(\Leftrightarrow a^2+2\ge2\sqrt{a^2+1}\)
\(\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\)
\(\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) ( LĐ )
Dấu "=" xảy ra \(\Leftrightarrow a=0\).
Bài 2: tương tự 1b.
Bài 3:
Do \(a,b,c\) dương nên ta có các BĐT:
\(\frac{a}{a+b+c}< \frac{a}{a+b}< \frac{a+c}{a+b+c}\)
Tương tự: \(\frac{b}{a+b+c}< \frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{a+b+c}< \frac{c}{c+a}< \frac{c+b}{a+b+c}\)
Cộng theo vế 3 BĐT:
\(\frac{a+b+c}{a+b+c}< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)( đpcm )
\(\sqrt{\frac{a}{c+b}}=\frac{a}{\sqrt{a\left(c+b\right)}}\ge\frac{a}{\frac{a+b+c}{2}}=\frac{2a}{a+b+c}\)
tương tự : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)(ĐPCM)
Ta có: \(\sqrt{\frac{a}{b+c}}=\sqrt{\frac{a^2}{a\left(b+c\right)}}\ge\frac{2a}{a+b+c}\)(1)
Tương tự: \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\)(2)
\(\sqrt{\frac{c}{b+a}}\ge\frac{2c}{a+b+c}\)(3)
Cộng (1),(2),(3) vế theo vế => \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge2\)
Đặt \(\left(\frac{a}{b};\sqrt{\frac{b}{c}};\sqrt[3]{\frac{c}{a}}\right)=\left(x;y;z\right)\Rightarrow xy^2z^3=1\)
\(P=x+y+z=x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3}\)
\(P\ge6\sqrt[6]{\frac{xy^2z^3}{108}}=\frac{6}{\sqrt[6]{108}}=\sqrt[6]{432}>\frac{5}{2}\)