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![](https://rs.olm.vn/images/avt/0.png?1311)
a) =
=
b) =
=
=
. ( 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) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{a^{\dfrac{1}{3}}\cdot\sqrt{b}+b^{\dfrac{1}{3}}\cdot\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}-\sqrt[3]{ab}\)
\(=\dfrac{a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{2}}+b^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{3}{6}}+a^{\dfrac{3}{6}}\cdot b^{\dfrac{2}{6}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{2}{6}}\left(a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}\right)}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sum_{sym}\sqrt{\dfrac{a^4+b^4}{1+ab}}=\sum_{sym}\sqrt{\dfrac{2\left(a^4+b^4\right)}{2+2ab}}>=\sum_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}+\sum_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}\)
\(\sum_{cyc}\dfrac{a^2}{\sqrt{2+2ab}}>=\dfrac{2\left(a+b+c\right)^2}{\sum2\sqrt{2+2ab}}>=\dfrac{3}{2}\)
\(\sum_{cyc}\dfrac{b^2}{\sqrt{2+2ab}}>=\dfrac{3}{2}\)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ca}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)
Áp dụng BĐT Caushy-Schwarz ta được:
\(\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}}\)
Ta chứng minh rằng:
\(a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}\le\dfrac{4}{3}\left(a+b+c\right)^2\)
\(\Leftrightarrow\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\dfrac{4}{3}\left(a+b+c\right)^2\)
Áp dụng BĐT Bunhiacopxki ta được:
\(\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+45abc\right)}\)Ta tiếp tục chứng minh:
\(\dfrac{16}{9}\left(a+b+c\right)^3\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge a^3+b^3+c^3+45abc\)
Áp dụng BĐT AM-GM (Caushy) ta được:
\(\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge\dfrac{16}{9}\left(a^3+b^3+c^3+3.2\sqrt{ab}.2.\sqrt{bc}.2.\sqrt{ca}\right)=\dfrac{16}{9}.\left(a^3+b^3+c^3+24abc\right)\)
Ta chứng minh:
\(\dfrac{16}{9}\left(a^3+b^3+c^3+24abc\right)\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{16}{9}a^3+\dfrac{16}{9}b^3+\dfrac{16}{9}c^3+\dfrac{16}{9}.24abc\ge a^3+b^3+c^3+45abc\)
\(\Leftrightarrow\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{3}abc\) (*)
Áp dụng BĐT AM-GM (Caushy) ta được:
\(\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{9}.3\sqrt[3]{a^3b^3c^3}=\dfrac{7}{3}abc\)
\(\Rightarrow\) (*) đúng.
Vậy BĐT đã được chứng minh. Dấu "=" xảy ra khi \(a=b=c>0\).
2.
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= ![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?b%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Câu a, b thì Nguyễn Quang Duy làm đúng rồi.
c) \(a^{\dfrac{4}{3}}:\sqrt[3]{a}=a^{\dfrac{4}{3}}:a^{\dfrac{1}{3}}=a^{\dfrac{4}{3}-\dfrac{1}{3}}=a\)
d) \(\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}\)