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\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
\(1.\) 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....
Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=z^2\\b^2+c^2=x^2\\c^2+a^2=y^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
Điều kiện đề bài thành: \(x+y+z=3\sqrt{2}\)
Ta có:
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
\(=\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\)
\(=\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-x-y-z\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-x-y-z\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2\sqrt{2}}.3\sqrt{2}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{2}\) hay \(a=b=c=1\)
bạn có cách làm kh ạ? Chỉ tớ =((