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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:
Áp dụng BĐT AM-GM ngược dấu ta có:
\(A=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ac}}=\frac{ab}{\sqrt{c(a+b+c)+ab}}+\frac{bc}{\sqrt{a(a+b+c)+bc}}+\frac{ca}{\sqrt{b(a+b+c)+ac}}\)
\(=\frac{ab}{\sqrt{(c+a)(c+b)}}+\frac{bc}{\sqrt{(a+b)(a+c)}}+\frac{ca}{\sqrt{(b+a)(b+c)}}\)
\(\leq \frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)+\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)+\frac{1}{2}\left(\frac{ca}{b+a}+\frac{ca}{b+c}\right)\)
\(A\leq \frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}(b+a+c)=\frac{1}{2}\)
Vậy \(A_{\max}=\frac{1}{2}\) tại \(a=b=c=\frac{1}{3}\)
\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)
\(\Leftrightarrow3< 1\) ( Vô lý )
\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)
\(\Leftrightarrow2b-2\sqrt{ab}< 0\)
\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)
Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)
\(\RightarrowĐpcm.\)
\(2a.\) Áp dụng BĐT Cauchy , ta có :
\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)
\(\Leftrightarrow x-4=a^2\)
\(\Leftrightarrow x=a^2+4\left(TM\right)\)
\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+4=x^2+4x+4\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
KL....
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Bài 3:
\(C=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a-1}{\sqrt{a}+3}\)
\(=\dfrac{\left(a-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
Ta có:
\(\dfrac{ab}{\sqrt{c+ab}}=\dfrac{ab}{\sqrt{c\left(a+b+c\right)+ab}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\dfrac{\sqrt{ab}}{\sqrt{a+c}}.\dfrac{\sqrt{ab}}{\sqrt{b+c}}\)
\(\Rightarrow\dfrac{ab}{\sqrt{c+ab}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta có:
\(\dfrac{bc}{\sqrt{a+bc}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\) ; \(\dfrac{ac}{\sqrt{b+ac}}\le\dfrac{1}{2}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)\)
Cộng vế với vế ta được:
\(A\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}\right)\)
\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right)\)
\(\Rightarrow A\le\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
\(\Rightarrow A_{max}=\dfrac{1}{2}\) khi \(a=b=c=\dfrac{1}{3}\)
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Áp dụng BĐT AM-GM:
\(\frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{b}{a+b}+\frac{c}{a+c})\)
Hoàn toàn tương tự:
\(\frac{1}{\sqrt{b^2+1}}=\sqrt{\frac{ac}{(b+a)(b+c)}}\leq \frac{1}{2}(\frac{a}{b+a}+\frac{c}{b+c})\)
\(\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{a}{c+a}+\frac{b}{b+c})\)
Cộng theo vế:
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\leq \frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
\(a+b+c+2=abc\)
\(\Leftrightarrow2a+2b+2c+3+ab+bc+ca=abc+ab+bc+ca+a+b+c+1\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)+\left(c+1\right)\left(b+1\right)+\left(c+1\right)\left(a+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=1\)
Đặt \(\left(\dfrac{1}{a+1};\dfrac{1}{b+1};\dfrac{1}{c+1}\right)=\left(x;y;z\right)\)
\(\Rightarrow x+y+z=1\)
BĐT trở thành:
\(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}=\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{3}\) hay \(a=b=c=2\)