Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)
Vì \(a=\left(a-1\right)+1\ge2\sqrt{\left(a-1\right).1}=2\sqrt{a-1}\)
\(b=\left(b-4\right)+4\ge2\sqrt{\left(b-4\right).4}=4\sqrt{b-4}\)
\(c=\left(c-9\right)+9\ge2\sqrt{\left(c-9\right).9}=6\sqrt{c-9}\)
=>\(P\le\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)
P max = 11/12 khi a=2; b=8; c =18
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 \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}=\frac{\sqrt{\left(a-1\right)\cdot1}}{a}+\frac{1}{2}\cdot\frac{\sqrt{\left(b-4\right)\cdot4}}{b}+\frac{1}{3}\cdot\frac{\sqrt{\left(c-9\right)\cdot9}}{c}\)
\(\Rightarrow P\le\frac{\frac{a-1+1}{2}}{a}+\frac{1}{2}\cdot\frac{\frac{b-4+4}{2}}{b}+\frac{1}{3}\cdot\frac{\frac{c-9+9}{2}}{c}\)
\(\Rightarrow P\le\frac{a}{2a}+\frac{b}{4b}+\frac{c}{6c}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=18\end{matrix}\right.\)
\(P=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-4}}{b}+\dfrac{\sqrt{c-9}}{c}=\dfrac{1.\sqrt{a-1}}{a}+\dfrac{2.\sqrt{b-4}}{2b}+\dfrac{3.\sqrt{c-9}}{3c}\)
Áp dụng hằng đẳng thức \(xy\le\dfrac{x^2+y^2}{2}\) ta được
\(P\le\dfrac{1+a-1}{2a}+\dfrac{4+b-4}{4b}+\dfrac{9+c-9}{6c}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{11}{12}\)
\(\Rightarrow P_{max}=\dfrac{11}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{a-1}=1\\\sqrt{b-4}=2\\\sqrt{c-9}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=18\end{matrix}\right.\)
*Sửa đề: tìm GTNN
\(A=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\)
\(=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\ge\frac{\frac{2+c-2}{2}}{\sqrt{2}c}=\frac{\frac{c}{2}}{\sqrt{2}c}=\frac{1}{2\sqrt{2}}\)
TƯơng tự cho 2 BĐT còn lại ta cũng có:
\(\frac{\sqrt{a-3}}{a}\ge\frac{1}{2\sqrt{3}};\frac{\sqrt{b-4}}{b}\ge\frac{1}{2\sqrt{4}}\)
Suy ra \(A\ge\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)\)