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a) =
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b) =
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. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
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Theo giả thiết kết hợp sử dụng BĐT AM - GM có:
\(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-\left[c\left(a+b\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)
\(\le\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-2\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\left[\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\right]^2\)
Suy ra \(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\ge2\Leftrightarrow\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}\ge3\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge7\)
Khi đó, sử dụng BĐT Cauchy - Schwarz ta có:
\(\left(a^4+b^4+c^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\ge\left[\sqrt{\left(a^4+b^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}\right)}+1\right]^2\)
\(=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+1\right)^2=\left[\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-1\right]^2\ge\left(7^2-1\right)^2=2304\)
Đẳng thức xảy ra khi và chỉ khi \(ab=c^2\) và \(\dfrac{a}{b}+\dfrac{b}{a}=7\)
(a+b-c)(1/a+1/b-c)=(a+b)(1/a+1/b)+1-[c(a+b)+c(1/a+1/b)]<=(a+b)(1/a+1/b)+1-2căn (a+b)(1/a+1/b)
=[(căn (a+b)(1/a+1/b))-1]^2
=>\(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1>=2\)
=>\(\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}>=3\)
=>a/b+b/a>=7
(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=[căn ((a^4+b^4)(1/a^4+1/b^4))+1]^2
=(a^2/b^2+b^2/a^2+1)^2=[(a/b+b/a)^2-1]^2>=(7^2-1)^2=2304
=>ĐPCM
\(\sum_{sym}\sqrt{\dfrac{a^4+b^4}{1+ab}}=\sum_{sym}\sqrt{\dfrac{2\left(a^4+b^4\right)}{2+2ab}}>=\sum_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}+\sum_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}\)
\(\sum_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}>=\dfrac{2\left(a+b+c\right)^2}{\sum2\sqrt{2+2ab}}>=\dfrac{3}{2}\)
\(\sum_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}>=\dfrac{3}{2}\)
Cộng các BĐT trên, ta được ĐPCM
Ta có:
\(\Sigma_{sym}\sqrt{\dfrac{a^4+b^4}{1+ab}}=\Sigma_{sym}\sqrt{\dfrac{2\left(a^4+b^4\right)}{2+2ab}}\ge\Sigma_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}+\Sigma_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}\)
Sử dụng BĐT Cauchy - Schwarz và AM - GM có:
\(\Sigma_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}\ge\dfrac{2\left(a+b+c\right)^2}{\Sigma2\sqrt{2+2ab}}\ge\dfrac{2\left(a+b+c\right)^2}{ab+bc+ca+9}\ge\dfrac{3}{2}\)
Tương tự: \(\Sigma_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}\ge\dfrac{3}{2}\)
Cộng 2 BĐT ta được:
\(\sqrt{\dfrac{a^4+b^4}{1+ab}}+\sqrt{\dfrac{b^4+c^4}{1+bc}}+\sqrt{\dfrac{c^4+a^4}{1+ca}}\ge3\)
Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)
Ta có:
\(\left(b-\dfrac{1}{2}\right)^2\ge0\) <=> \(b^2-b+\dfrac{1}{4}\ge0\) <=>\(b-\dfrac{1}{4}\le b^2\)
Mà :
a<1 => \(log_a\left(b-\dfrac{1}{4}\right)\ge log_ab^2=2log_ab\)
P=\(log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}log_{\dfrac{a}{b}}b=log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\ge2log_ab-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\)
Đặt t=logab
Do b<a<1 => t=logab >1
Khi đó \(P\ge2t+\dfrac{t}{2t-2}=f\left(t\right)\). Khảo sát f(t) trên (1;+\(\infty\)) ta đc
P\(\ge\)f(t) \(\ge\) f\(\left(\dfrac{3}{2}\right)\) = \(\dfrac{9}{2}\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
\(P=\frac{1}{2}log_{\frac{a}{b}}a-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2log_a\frac{a}{b}}-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2\left(1-log_ab\right)}-4log_a\left(a+\frac{b}{4}\right)\)
Ta có: \(a+\frac{b}{4}\ge2\sqrt{\frac{ab}{4}}=\sqrt{ab}\)
\(\Rightarrow log_a\left(a+\frac{b}{4}\right)\le log_a\sqrt{ab}\) (do \(0< a< 1\))
\(\Rightarrow P\ge\frac{1}{2\left(1-log_ab\right)}-4log_a\sqrt{ab}=\frac{1}{2\left(1-log_ab\right)}-2\left(1+log_ab\right)\)
Đặt \(log_ab=x\Rightarrow0< x< 1\) \(\Rightarrow P\ge\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\)
Xét hàm \(f\left(x\right)=\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\) với \(0< x< 1\)
\(f'\left(x\right)=\frac{1}{2\left(1-x\right)^2}-2=0\Leftrightarrow\frac{1-4\left(1-x\right)^2}{2\left(1-x\right)^2}=0\Rightarrow x=\frac{1}{2}\)
Từ BBT ta thấy \(f\left(x\right)_{min}=f\left(\frac{1}{2}\right)=-2\)
\(\Rightarrow P\ge-2\Rightarrow P_{min}=-2\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}log_ab=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\a=\frac{b}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{16}\\b=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow S=\frac{5}{16}\)
Chọn B