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\(\Leftrightarrow y=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)
Ta có: \(\dfrac{\sqrt{c-2}}{c}\le\dfrac{1}{2\sqrt{2}}\Leftrightarrow\left(\sqrt{c-2}-\sqrt{2}\right)^2\ge0\) ( Luôn đúng)
Tương tự: \(\dfrac{\sqrt{a-3}}{a}\le\dfrac{1}{2\sqrt{3}};\dfrac{\sqrt{b-4}}{b}\le\dfrac{1}{4}\)
\(\Rightarrow y\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{4}\) và dấu ''='' xảy ra khi c = 4; a = 6; b = 8
*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 \(A\ge\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)\)
\(\left(\frac{1}{2\sqrt{2}}c-\sqrt{2}\right)^2\ge0\)
\(\Rightarrow\frac{1}{8}c^2-c+2\ge0\)
\(\Rightarrow\frac{1}{2\sqrt{2}}c\ge\sqrt{c-2}\)
\(\Rightarrow\frac{1}{2\sqrt{2}}\ge\frac{\sqrt{c-2}}{c}\)
tương tự \(\left(\frac{1}{2\sqrt{3}}a-\sqrt{3}\right)^2\ge0\Rightarrow\frac{1}{2\sqrt{3}}\ge\frac{\sqrt{a-3}}{a}\)
\(\left(\frac{1}{4}b-2\right)^2\ge0\Rightarrow\frac{1}{4}\ge\frac{\sqrt{b-4}}{b}\)
\(\Rightarrow P\le\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)
Dấu "=" xảy ra <=>c=4;a=6;b=8
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Ta có : \(\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ac\sqrt{b-4}}{abc}=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng bất đẳng thức Cauchy, ta có :
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\le\frac{2+c-2}{2\sqrt{2}c}=\frac{1}{2\sqrt{2}}\)
\(\frac{\sqrt{a-3}}{a}=\frac{\sqrt{3\left(a-3\right)}}{\sqrt{3}a}\le\frac{3+a-3}{2\sqrt{3}a}=\frac{1}{2\sqrt{3}}\)
\(\frac{\sqrt{b-4}}{b}=\frac{\sqrt{4\left(b-4\right)}}{2b}\le\frac{4+b-4}{4b}=\frac{1}{4}\)
\(\Rightarrow\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\le\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}c-2=2\\b-4=4\\a-3=3\end{cases}\Leftrightarrow}\hept{\begin{cases}c=4\\b=8\\a=6\end{cases}}\)
Vậy giá trị lớn nhất của biểu thức là \(\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\Leftrightarrow\hept{\begin{cases}a=6\\b=8\\c=4\end{cases}}\)
phá ra nha
sau đó bạn lm theo tek này
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\le\frac{\frac{c}{2}}{\sqrt{2}c}=\frac{1}{\sqrt{2}}\)
mấy cái kia tt nha
Lời giải:
Do \(a\geq 1; b\geq 2; c\geq 3\Rightarrow a-1, b-2, c-3\geq 0\)
Áp dụng BĐT AM-GM cho các số không âm ta có:
\(\left\{\begin{matrix} (a-1)+4\geq 2\sqrt{4(a-1)}=4\sqrt{a-1}\\ (b-2)+9\geq 2\sqrt{9(b-2)}=6\sqrt{b-2}\\ (c-3)+16\geq 2\sqrt{16(c-3)}=8\sqrt{c-3}\end{matrix}\right.\)
Cộng theo vế và rút gọn thu được:
\(a+b+c+23\geq 4\sqrt{a-1}+6\sqrt{b-2}+8\sqrt{c-3}\) (đpcm)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} a-1=4\\ b-2=9\\ c-3=16\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=5\\ b=11\\ c=19\end{matrix}\right.\)
biến đổi ta đc \(P=\dfrac{\sqrt{c-1}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-2}}{b}\)
ta có \(c=c-1+1\ge2\sqrt{c-1}\)
=> \(\dfrac{\sqrt{c-1}}{c}\le\dfrac{1}{2}\)
tương tự ta có \(\dfrac{\sqrt{b-2}}{b}\le\dfrac{1}{2\sqrt{2}}\); \(\dfrac{\sqrt{a-3}}{a}\le\dfrac{1}{2\sqrt{3}}\)
=> P \(\le\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\)
dấu đẳng thức xảy ra khi c=2;b=4;a=6