Ta có:

\(bc\sqrt{1\left(a-1\right)}\le bc.\frac{1+a-1}{2}=\frac{abc}{2}\)

\(ca\sqrt{b-4}=\frac{1}{2}ca\sqrt{4\left(b-4\right)}\le\frac{1}{2}ca.\frac{4+b-4}{2}=\frac{abc}{4}\)

\(ab\sqrt{c-9}=\frac{1}{3}ab.\sqrt{9\left(c-9\right)}\le\frac{1}{3}ab.\frac{9+c-9}{2}=\frac{abc}{6}\)

Từ đó suy ra \(P\le\frac{abc\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right)}{abc}=\frac{11}{12}\)

Đẳng thức xảy ra khi a = 2; b = 8; c = 18

Is that true?

ta có P=\(\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)

Áp dụng bđt cố si ta có 

\(\sqrt{a-1}\le\frac{1}{2}\left(a-1+1\right)=\frac{1}{2}a\Rightarrow\frac{\sqrt{a-1}}{a}\le\frac{1}{2}\)

Tương tự mấy cái kia rồi + vào, để ý dấu = 

Bạn tham khảo tại đây ạ!

Câu hỏi của danh Vô - Toán lớp 9 - Học toán với OnlineMath

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\(=\frac{3}{2}\Leftrightarrow a=2;b=8;c=18\)

\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}=\frac{\sqrt{\left(a-1\right)\cdot1}}{a}+\frac{1}{2}\cdot\frac{\sqrt{\left(b-4\right)\cdot4}}{b}+\frac{1}{3}\cdot\frac{\sqrt{\left(c-9\right)\cdot9}}{c}\)

\(\Rightarrow P\le\frac{\frac{a-1+1}{2}}{a}+\frac{1}{2}\cdot\frac{\frac{b-4+4}{2}}{b}+\frac{1}{3}\cdot\frac{\frac{c-9+9}{2}}{c}\)

\(\Rightarrow P\le\frac{a}{2a}+\frac{b}{4b}+\frac{c}{6c}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=18\end{matrix}\right.\)

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

Ta có:

\(\dfrac{P}{1152}=\dfrac{bc\sqrt{a-1}+ca\sqrt{b-9}+ab\sqrt{c-16}}{1152}=\dfrac{bc\sqrt{a-1}+ca\sqrt{b-9}+ab\sqrt{c-16}}{abc}\)

\(\Leftrightarrow\dfrac{P}{1152}=\dfrac{1.\sqrt{a-1}}{a}+\dfrac{3.\sqrt{b-9}}{3b}+\dfrac{4\sqrt{c-16}}{4c}\)

\(\Rightarrow\dfrac{P}{1152}\le\dfrac{1+a-1}{2a}+\dfrac{9+b-9}{6b}+\dfrac{16+c-16}{8c}=\dfrac{19}{24}\)

\(\Rightarrow P\le912\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;18;36\right)\)

Chắc chắn rằng đề bài thiếu, biểu thức này ko tồn tại max

\(P=bc.1.\sqrt{a-1}+\dfrac{ca}{3}.3.\sqrt{b-9}+\dfrac{ab}{4}.4.\sqrt{c-16}\)

\(P\le\dfrac{bc}{2}\left(1+a-1\right)+\dfrac{ca}{6}\left(9+b-9\right)+\dfrac{ab}{8}\left(16+c-16\right)\)

\(\Rightarrow P\le\dfrac{abc}{2}+\dfrac{abc}{6}+\dfrac{abc}{8}=912\)

\(P_{max}=912\)  khi \(\left(a;b;c\right)=\left(2;18;32\right)\)