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a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
2. \(BĐT\Leftrightarrow\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\ge1\)
Đặt\(\frac{2}{a}=x;\frac{2}{b}=y;\frac{2}{c}=z\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=8\end{cases}}\)
Ta cần chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge1\Leftrightarrow\left(yz+y+z+1\right)+\left(zx+z+x+1\right)+\left(xy+x+y+1\right)\ge xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\)\(\Leftrightarrow x+y+z\ge6\)(Đúng vì \(x+y+z\ge3\sqrt[3]{xyz}=6\))
Đẳng thức xảy ra khi x = y = z = 2 hay a = b = c = 1
3. Ta có: \(a+b+c\le\sqrt{3}\Rightarrow\left(a+b+c\right)^2\le3\)
Ta có đánh giá quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Từ đó suy ra \(ab+bc+ca\le1\)
\(A=\frac{\sqrt{a^2+1}}{b+c}+\frac{\sqrt{b^2+1}}{c+a}+\frac{\sqrt{c^2+1}}{a+b}\ge\frac{\sqrt{a^2+ab+bc+ca}}{b+c}+\frac{\sqrt{b^2+ab+bc+ca}}{c+a}+\frac{\sqrt{c^2+ab+bc+ca}}{a+b}\)\(=\frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}+\frac{\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}+\frac{\sqrt{\left(c+a\right)\left(c+b\right)}}{a+b}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=3\)Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
3, \(P=a+b+\frac{1}{2a}+\frac{2}{b}\)
=\(\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\)
AD bđt cosi vs hai số dương có:
\(\frac{1}{2a}+\frac{a}{2}\ge2\sqrt{\frac{1}{2a}.\frac{a}{2}}=2\sqrt{\frac{1}{4}}=1\)
\(\frac{b}{2}+\frac{2}{b}\ge2\sqrt{\frac{b}{2}.\frac{2}{b}}=2\)
Có \(\frac{a+b}{2}\ge\frac{3}{2}\) (vì a+b \(\ge3\))
=> \(P=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\ge1+2+\frac{3}{2}\)
<=> P \(\ge4.5\)
Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}\frac{1}{2a}=\frac{a}{2}\\\frac{b}{2}=\frac{2}{b}\\a+b=3\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a^2=1\\b^2=4\\a+b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=1\\b=2\\a+b=3\end{matrix}\right.\)
=> a=2,b=3
Vậy minP=4.5 <=>a=1,b=2
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{16\left(a+b+c\right)^2}+\frac{1215}{16\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\frac{2.9}{4}+\frac{1215.4}{16.9}}=\frac{3\sqrt{17}}{2}\)
√a2+1b2 +√b2+1c2 +√c2+1a2
≥√(a+b+c)2+(1a +1b +1c )2
≥√(a+b+c)2+81(a+b+c)2
≥√(a+b+c)2+8116(a+b+c)2 +121516(a+b+c)2
≥√2.94 +1215.416.9 =3√172
Em làm thử nhé!
Bài 1: \(A=\left[\frac{a^2}{b-1}+4\left(b-1\right)\right]+\left[\frac{b^2}{a-1}+4\left(a-1\right)\right]-4\left(a+b\right)+8\)
Cauchy vào là ra rồi ạ;)
Bài 2: Em chịu
2) Có: \(\sqrt{ab}\le\frac{a+b}{2}=1\); \(\sqrt{a}+\sqrt{b}=\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}\le\sqrt{2\left(a+b\right)}=2\)
\(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}\ge\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3=\frac{a^2}{\sqrt{a}}+\frac{b^2}{\sqrt{b}}\)
\(\ge\frac{\left(a+b\right)^2}{\sqrt{a}+\sqrt{b}}\ge=\frac{2^2}{2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=1\)
b2 \(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=\sqrt{x}.\sqrt{1-\frac{1}{x}}+\sqrt{y}.\)\(\sqrt{y}.\sqrt{1-\frac{1}{y}}+\sqrt{z}.\sqrt{1-\frac{1}{z}}\)rồi dung bunhia là xong
A= \(\frac{1}{a^3}\)+ \(\frac{1}{b^3}\)+ \(\frac{1}{c^3}\)+ \(\frac{ab^2}{c^3}\)+ \(\frac{bc^2}{a^3}\)+ \(\frac{ca^2}{b^3}\)
Svacxo:
3 cái đầu >= \(\frac{9}{a^3+b^3+c^3}\)
3 cái sau >= \(\frac{\left(\sqrt{a}b+\sqrt{c}b+\sqrt{a}c\right)^2}{a^3+b^3+c^3}\)
Cô-si: cái tử bỏ bình phương >= 3\(\sqrt{abc}\)
=> cái tử >= 9abc= 9 vì abc=1
Còn lại tự làm
\(S\ge3\sqrt[6]{\frac{a^2b^2+1}{ab}.\frac{b^2c^2+1}{bc}.\frac{c^2a^2+1}{ca}}\)