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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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Chắc đề bị nhầm rồi.
\(\dfrac{a}{\sqrt{b+1}}+\dfrac{b}{\sqrt{c+1}}+\dfrac{c}{\sqrt{a+1}}\ge2\sqrt{2}\left(\dfrac{a}{3+b}+\dfrac{b}{3+c}+\dfrac{c}{3+a}\right)\)
\(\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(ab+bc+ca\right)}\ge2\sqrt{2}.\dfrac{9}{9+\dfrac{\left(a+b+c\right)^2}{3}}=2\sqrt{2}.\dfrac{9}{12}=\dfrac{3}{\sqrt{2}}\)
\(1.\) 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....
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(M^2=(a\sqrt{9b(a+8b)}+b\sqrt{9a(b+8a)})^2\)
\(\leq (a^2+b^2)(9ab+72b^2+9ab+72a^2)\)
\(\Leftrightarrow M^2\leq (a^2+b^2)(72a^2+72b^2+18ab)\)
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 18ab\leq 9(a^2+b^2)\)
Do đó, \(M^2\leq (a^2+b^2)(72a^2+72b^2+9a^2+9b^2)=81(a^2+b^2)^2\)
\(\Leftrightarrow M\leq 9(a^2+b^2)\leq 144\)
Vậy \(M_{\max}=144\Leftrightarrow a=b=\sqrt{8}\)
Bài 6:
\(a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\)
Vì \(a>1\rightarrow a-1>0\). Do đó áp dụng BĐT Am-Gm cho số dương\(a-1,\frac{1}{a-1}\) ta có:
\((a-1)+\frac{1}{a-1}\geq 2\sqrt{\frac{a-1}{a-1}}=2\)
\(\Rightarrow a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\geq 3\) (đpcm)
Dấu bằng xảy ra khi \(a-1=1\Leftrightarrow a=2\)
Bài 3:
Xét \(\sqrt{a^2+1}\). Vì \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(\Rightarrow \sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM có: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)
hay \(\sqrt{a^2+1}\leq \frac{2a+b+c}{2}\)
Hoàn toàn tương tự với các biểu thức còn lại và cộng theo vế:
\(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\leq \frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}=2(a+b+c)\)
Ta có đpcm. Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bài 4:
Ta có:
\(A=\frac{8a^2+b}{4a}+b^2=2a+\frac{b}{4a}+b^2\)
\(\Leftrightarrow A+\frac{1}{4}=2a+\frac{b+a}{4a}+b^2=2a+b+\frac{b+a}{4a}+b^2-b\)
Vì \(a+b\geq 1, a>0\) nên \(A+\frac{1}{4}\geq a+1+\frac{1}{4a}+b^2-b\)
Áp dụng BĐT AM-GM:
\(a+\frac{1}{4a}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\Rightarrow A+\frac{1}{4}\geq 2+b^2-b=\left(b-\frac{1}{2}\right)^2+\frac{7}{4}\geq \frac{7}{4}\)
\(\Leftrightarrow A\geq \frac{3}{2}\).
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=\frac{1}{2}\)
Câu a
\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right):\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{1}\)
\(=a-b\)
a/ \(\sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=3\sqrt{9a^2}=3.3.a=9a\) ( vì \(a\ge0\) )
b/ \(\sqrt{8a}.\sqrt{\dfrac{32}{a}}=2\sqrt{2a}.\dfrac{4\sqrt{2a}}{\sqrt{a}}=2\sqrt{2}.4\sqrt{2a}=8\sqrt{4a}=16\sqrt{a}\)
c/ĐK: x\(\ge\)0
\(\sqrt{2x}.\sqrt{3x}.\sqrt{6x^2}=\sqrt{2x}.\sqrt{3x}.\left|x\right|\sqrt{2}.\sqrt{3}\) = \(x.2\sqrt{x}.3\sqrt{x}\) ( vì x\(\ge0\) ) = \(6x^2\)
d/ Đk: x>0
\(\sqrt{3x}.\sqrt{\dfrac{48}{x}}=\sqrt{3}.\sqrt{x}.\dfrac{4\sqrt{3}}{\sqrt{x}}=4.3=12\)
e/ \(\dfrac{1}{a-1}.\sqrt{9\left(a-1\right)^2}=\dfrac{1}{a-1}.\sqrt{9}.\sqrt{\left(a-1\right)^2}\)
= \(\dfrac{1}{a-1}.3.\left|a-1\right|=\dfrac{1}{a-1}.3.\left(a-1\right)\) ( vì a>1)
= 3
Đặt \(a^{\dfrac{1}{9}};b^{\dfrac{1}{9}};c^{\dfrac{1}{9}}\rightarrow x;y;z\)\(\left(x;y;z>0;xyz=1\right)\)
Ta có BĐT:\(\dfrac{1}{\sqrt{8x^9+1}}\ge\dfrac{1}{x^8+x^4+1}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(x-1\right)^2x^4\left(x^{10}+2x^9+3x^8+4x^7+7x^6+10x^5+13x^4+8x^3+6x^2+4x+2\right)}{\left(x^2-x+1\right)^2\left(x^2+x+1\right)^2\left(2x^3+1\right)\left(x^4-x^2+1\right)^2\left(4x^6-2x^3+1\right)}}{\dfrac{1}{\sqrt{8x^9+1}}+\dfrac{1}{x^8+x^4+1}}\ge0\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(A\ge\dfrac{1}{x^8+x^4+1}+\dfrac{1}{y^8+y^4+1}+\dfrac{1}{z^8+z^4+1}\ge1\)
Dấu "=" khi \(x=y=z=a=b=c=1\)
đặt a,b,c=1/9 hả bạn