Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét a=1,b=4,c=9 thì P=0
Xét \(a>1,b>4,c>9\)
Áp dụng BĐT AM-GM ta có:
\(P=\frac{bc.\sqrt{a-1}.1+\frac{ca}{2}.\sqrt{b-4}.2+\frac{ab}{3}.\sqrt{c-9}.3}{abc}\)
\(\le\frac{bc.\frac{a-1+1}{2}+\frac{ca}{2}.\frac{b-4+4}{2}+\frac{ab}{3}.\frac{c-9+9}{2}}{abc}\)
\(=\frac{\frac{abc}{2}+\frac{abc}{4}+\frac{abc}{6}}{abc}=\frac{\frac{11}{12}abc}{abc}=\frac{11}{12}\)
Nên GTLN của P là \(\frac{11}{12}\) đạt được khi \(\hept{\begin{cases}\sqrt{a-1}=1\\\sqrt{b-4}=2\\\sqrt{c-9}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}a-1=1\\b-4=4\\c-9=9\end{cases}\Leftrightarrow}\hept{\begin{cases}a=2\\b=8\\c=18\end{cases}}\)
\(P=\frac{bc\sqrt{a-1}+ca\sqrt{b-4}+ab\sqrt{c-9}}{abc}=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)
Vì \(a\ge1;b\ge4;c\ge9\). Áp dụng BĐT Cosi cho các số dương ta được:
\(\sqrt{a-1}=1\cdot\sqrt{a-1}\le\frac{1+a-1}{2}=\frac{a}{2}\). Dấu "=" xảy ra \(\Leftrightarrow\sqrt{a-1}=1\Leftrightarrow a=2\)
\(\sqrt{b-4}=2\cdot\sqrt{b-4}\le\frac{4+b-4}{2}=\frac{b}{2}\). Dấu "=" xảy ra \(\Leftrightarrow\sqrt{b-4}=2\Leftrightarrow b=8\)
\(\sqrt{c-9}=3\cdot\sqrt{c-9}\le\frac{9+c-9}{2}=\frac{c}{2}\). Dấu "=" xảy ra \(\Leftrightarrow\sqrt{c-9}=3\Leftrightarrow c=18\)
\(\Rightarrow P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\le\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}=\frac{3}{2}\)
Vậy GTLN của P\(=\frac{3}{2}\Leftrightarrow a=2;b=8;c=18\)
Áp dụng bất đẳng thức Cô-si, ta được: \(P=\frac{bc\sqrt{a-1}+ca\sqrt{b-4}+ab\sqrt{c-9}}{abc}\)\(=\frac{bc\sqrt{\left(a-1\right).1}+\frac{1}{2}ca\sqrt{4.\left(b-4\right)}+\frac{1}{3}ab\sqrt{9.\left(c-9\right)}}{abc}\)\(\le\frac{bc.\frac{\left(a-1\right)+1}{2}+\frac{1}{2}ca.\frac{4+\left(b-4\right)}{2}+\frac{1}{3}ab.\frac{9+\left(c-9\right)}{2}}{abc}\)\(=\frac{\frac{1}{2}abc+\frac{1}{4}abc+\frac{1}{6}abc}{abc}=\frac{\frac{11}{12}abc}{abc}=\frac{11}{12}\)
Đẳng thức xảy ra khi a = 2; b = 8; c = 18
\(P=bc\sqrt{a-1}+ca\sqrt{b-9}+ab\sqrt{c-16}\\ \Leftrightarrow\dfrac{P}{abc}=\dfrac{P}{1152}=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-9}}{b}+\dfrac{\sqrt{c-16}}{c}\)
Áp dụng BĐT Cauchy:
\(2\sqrt{a-1}\le a-1+1=a\Leftrightarrow\dfrac{\sqrt{a-1}}{a}\le\dfrac{1}{2}\\ 2\sqrt{9\left(b-9\right)}\le9+b-9=b\Leftrightarrow\dfrac{\sqrt{b-9}}{b}\le\dfrac{1}{6}\\ 2\sqrt{16\left(c-16\right)}\le16+b-16=c\Leftrightarrow\dfrac{\sqrt{c-16}}{c}\le\dfrac{1}{8}\)
Cộng VTV \(\Leftrightarrow\dfrac{P}{1152}\le\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}=\dfrac{19}{24}\)
\(\Leftrightarrow P\le912\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=1\\b-9=9\\c-16=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=18\\c=32\end{matrix}\right.\)
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
\(c+ab=\left(a+b+c\right)c+ab=ac+cb+c^2+ab=\left(a+c\right)\left(b+c\right)\)
Tương tự : \(a+bc=\left(a+b\right)\left(a+c\right);c+ab=\left(c+a\right)\left(c+b\right)\)
\(P=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)
áp dụng bất đẳng tức cauchy :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
cộng vế theo vế
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}+\frac{a}{b+a}\right)\)
\(\Leftrightarrow P\le\frac{1}{2}\left(\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{1}{2}\cdot3=\frac{3}{2}\)
dấu "=" xảy ra khi a=b=c=1/3
Có a+b+c=1 => c=(a+b+c).c=ac+bc+c2
\(\Rightarrow c+ab=ac+bc+c^2+ab=a\left(b+c\right)+c\left(b+c\right)=\left(b+c\right)\left(a+c\right)\)
\(\Rightarrow\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{\frac{a}{c+b}+\frac{b}{c+b}}{2}\)
Tương tự ta có \(\hept{\begin{cases}a+bc=\left(a+b\right)\left(a+c\right)\\b+ac=\left(b+a\right)\left(b+c\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{b+ca}}=\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\end{cases}}}\)
\(\Rightarrow P\le\frac{\frac{b}{a+b}+\frac{c}{c+a}+\frac{c}{b+c}+\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{c+b}}{2}\)\(=\frac{\frac{a+c}{a+c}+\frac{c+b}{c+b}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a, b, c, d là các chữ số thỏa mãn: ab+ca=da ab-ca=a Tìm giá trị của d.
\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)
Ta có: \(a=\left(a-1\right)+1\ge2\sqrt{a-1}\)
\(b=\left(b-4\right)+4\ge2\sqrt{\left(b-4\right).4}=4\sqrt{b-4}\)
\(c=\left(c-9\right)+9\ge2\sqrt{\left(c-9\right).9}=6\sqrt{c-9}\)
\(\Rightarrow P\le\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)
Dấu "=" xảy ra khi và chỉ khi \(a-1=1;b-4=4;c-9=9\)hay \(a=2;b=8;c=18\)