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Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)
\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)
\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)
Tương tự:
$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$
Khi đó:
\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)
\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)
\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)
\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)
Xét \(\sqrt{\dfrac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\dfrac{\left(a\left(a+b+c\right)+bc\right)\left(b\left(a+b+c\right)+ac\right)}{c\left(a+b+c\right)+ab}}\)
\(=\sqrt{\dfrac{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}{ac+bc+c^2+ab}}\)
\(=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}}\)\(=\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế
\(P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Ta có: bc(a2+1) = (a+b)(a+c)
\(\Rightarrow\) \(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}\) =\(\sqrt{\dfrac{a}{a+b}}.\sqrt{\dfrac{a}{a+c}}\)
Áp dụng BĐT Cô-si: \(\sqrt{\dfrac{a}{a+b}}.\sqrt{\dfrac{a}{a+c}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+c}+\dfrac{a}{a+c}\right)\)
\(\Rightarrow\) \(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
CMTT: \(\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{a+c}\right)\)
\(\dfrac{c}{\sqrt{ab\left(1+c^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\) S \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+a}+\dfrac{a}{c+a}+\dfrac{b}{a+b}+\dfrac{b}{c+b}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
\(\Rightarrow\) S\(\le\) \(\dfrac{1}{2}.3=\dfrac{3}{2}\)
Vậy Smax = \(\dfrac{3}{2}\)
Dấu "=" xảy ra\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=b=c\\a+b+c=abc\end{matrix}\right.\)
\(\Leftrightarrow\) \(a=b=c=\sqrt{3}\)
Lời giải:
Biểu thức có GTLN chứ không có GTNN bạn nhé. Nếu tìm GTLN thì làm như sau:
\(a+b+c=abc\)
\(\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=a^2bc+bc\)
\(\Rightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Rightarrow \frac{a}{\sqrt{bc(1+a^2)}}=\frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\) (theo BĐT AM-GM)
Hoàn toàn tương tự với các phân thức còn lại:
\(\frac{b}{\sqrt{ca(1+b^2)}}\leq \frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right);\frac{c}{\sqrt{ab(1+c^2)}}\leq \frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
Cộng theo vế các BĐT trên và rút gọn:
\(\Rightarrow \frac{a}{\sqrt{bc(1+a^2)}}+\frac{b}{\sqrt{ca(1+b^2)}}+\frac{c}{\sqrt{ab(1+c^2)}}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy GTLN là $\frac{3}{2}$. Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
Ta có:
\(\frac{2}{\sqrt{a}}+\frac{2}{\sqrt{b}}+\frac{2}{\sqrt{c}}=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)+\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\left(\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}\right)\)
\(\ge\frac{\left(1+1\right)^2}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+1\right)^2}{\sqrt{b}+\sqrt{c}}+\frac{\left(1+1\right)^2}{\sqrt{c}+\sqrt{a}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{4}{\sqrt{b}+\sqrt{c}}+\frac{4}{\sqrt{c}+\sqrt{a}}\)
=> \(2\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)\(\ge4\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
=> \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)\(\ge2\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
"=" xảy ra <=> a =b =c.
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Em nghĩ đề phải là \(\sqrt{c\left(a-c\right)}-\sqrt{c\left(b-c\right)}-\sqrt{ab}< 0\) chứ? Và em cũng không chắc đâu. Em mới biết sơ sơ về BĐT thôi. Nên nếu sai thì thông cảm cho em ạ
Từ đề bài suy ra \(\left(a-c\right)\left(b-c\right)>0\Leftrightarrow ab>ac+bc-c^2\)
Lại có \(ab>c\left(a-c\right)+bc>c\left(a-c\right)\) (do b và c không âm)
Suy ra \(\sqrt{c\left(a-c\right)}< \sqrt{ab}\)(1). Lại có: \(ab>ac+\left(bc-c^2\right)\)
\(=ac+c\left(b-c\right)>c\left(b-c\right)\Rightarrow\sqrt{c\left(b-c\right)}< \sqrt{ab}\) (2)
Từ (1) và (2) suy ra \(VT< \sqrt{ab}-\sqrt{ab}-\sqrt{ab}=-\sqrt{ab}\le0\)
Do vậy VT < 0 ta có đpcm.
a) \(BĐT\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
\(\Leftrightarrow\sqrt{\frac{c\left(a-c\right)}{ab}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)
\(\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)
Áp dụng AM-GM:\(VT\le\frac{1}{2}\left(\frac{c}{b}+1-\frac{c}{a}+\frac{c}{a}+1-\frac{c}{b}\right)=1\left(đpcm\right)\)
Dấu = xảy ra khi (a+b).c=ab
b) \(2+b+c+2+b+c\ge2\sqrt{\left(b+1\right)\left(c+1\right)}+2+b+c=\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge4\left(1+a\right)\)
\(\Leftrightarrow b+c\ge2a\)
cau a) dung cosi
\(\sqrt{c\left(a-c\right)}\le\frac{a-c+c}{2}\) ap dung cosi cho hai so c va a-c
tuong tu voi cac so khac
\(BT\le\frac{a-c+c}{2}+\frac{b-c+c}{2}-\frac{a+b}{2}\)(bt la VT cua de)
=> DPCM
b)
dung cosi nhu cau a
lam nhanh luon
\(\sqrt{1+b}\ge\frac{b+1+1}{2}\)
tuong tu
\(BT\ge\frac{b+2}{2}+\frac{c+2}{2}\ge a+2\)
<=> b+c>=2a