\(\frac{a^3}{\sqrt{b^2+3}}+\frac{a^3}{\sqrt{b^2+3}}+\frac{b^2+3}{8}\ge\frac{3}{2}a^2\)\(\Leftrightarrow\)\(\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}a^2-\frac{1}{16}b^2-\frac{3}{16}\)

\(P=\Sigma\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}\left(a^2+b^2+c^2\right)-\frac{1}{16}\left(a^2+b^2+c^2\right)-\frac{9}{16}=\frac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1 

different way

Ta co:

\(\text{ }P=\Sigma_{cyc}\frac{a^3}{\sqrt{b^2+3}}\ge\Sigma_{cyc}\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}a\sqrt{b^2+3}}\ge\frac{9}{\sqrt{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2+9\right)}}=\frac{3}{2}\)

Dau '=' xay ra khi \(a=b=c=1\)

(-5)+(+2)+I3I+(-4)+I-1I

=(-5)+2+3-4+1

=-3

Hok tốt nhé bạn ^^

#Dũng#

Chuyển 3x sang rồi bình phương hai vế nhớ đk bổ sung.

Đáp án : \(x=\frac{7-2\sqrt{10}}{9}\)

Đặt: \(\sqrt[3]{3x-1}=a;\sqrt[3]{4x-1}=b\)

\(\Rightarrow\sqrt[3]{12x^2-7x+1}=\sqrt[3]{\left(3x-1\right)\left(4x-1\right)}=ab\)

Phương trình có dạng :

 \(2a^2+3b^2=5ab\Leftrightarrow2a^2-5ab+3b^2=0\)

\(\Leftrightarrow2a^2-2ab-3ab+3b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\2a=3b\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt[3]{3x-1}=\sqrt[3]{4x-1}\\2\sqrt[3]{3x-1}=3\sqrt[3]{4x-1}\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=4x-1\\8\left(3x-1\right)=27\left(4x-1\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{19}{84}\end{cases}}}\)

Có: \(\frac{a^4}{b^2c}+\frac{b^4}{c^2a}+b\ge\frac{3ab}{c}\)

Tương tự, ta cũng được: \(\Sigma_{cyc}\frac{a^4}{b^2c}\ge\frac{3}{2}\Sigma_{cyc}\frac{ab}{c}-\frac{1}{2}\Sigma_{cyc}a\)

Cần CM: \(\Sigma_{cyc}\frac{ab}{c}\ge\Sigma_{cyc}a\)

Có: \(\frac{ab}{c}+\frac{bc}{a}\ge2b\)

Tương tự, ta có đpcm 

Dấu "=" xảy ra khi a=b=c