bố 32 tuổi

con 6 tuổi

ủng hộ nha

Câu b). Theo đầu bài ta có:
\(2a^2+2b^2=5ab\)
\(\Rightarrow2a^2+2b^2=ab+4ab\)
\(\Rightarrow2a^2+2b^2-4ab=ab\)
\(\Rightarrow2\left(a^2+b^2-2ab\right)=ab\)
\(\Rightarrow\left(a-b\right)^2=\frac{ab}{2}\)
\(\Rightarrow a-b=\sqrt{\frac{ab}{2}}\)
Mà \(2a^2+2b^2=5ab\)
\(\Rightarrow2a^2+2b^2=9ab-4ab\)
\(\Rightarrow2a^2+2b^2+4ab=9ab\)
\(\Rightarrow2\left(a^2+b^2+2ab\right)=9ab\)
\(\Rightarrow\left(a+b\right)^2=\frac{9ab}{2}\)
\(\Rightarrow a+b=\sqrt{\frac{9ab}{2}}\)
Từ trên suy ra:
\(Q=\frac{a+b}{a-b}=\left(a+b\right):\left(a-b\right)\)
\(\Leftrightarrow Q=\sqrt{\frac{9ab}{2}}:\sqrt{\frac{ab}{2}}\)
\(\Leftrightarrow Q=\sqrt{\frac{9ab}{2}:\frac{ab}{2}}\)
\(\Leftrightarrow Q=\sqrt{\frac{9\cdot ab\cdot2}{ab\cdot2}}\)
\(\Leftrightarrow Q=\sqrt{9}=3\)

Ta có: \(\frac{2a^2+3b^2}{2a^3+3b^3}\left(a+b\right)=1+ab\frac{2a+3b}{2a^3+3b^3}\)

Áp dụng BĐT Holder ta có: 

\(\left(2a^3+3b^3\right)\left(2+3\right)^2\ge\left(2a+3b\right)^3\)

Vậy ta có thể viết lại BĐT cần chứng minh như sau;

\(VT\left(a+b\right)\le2+25ab\left(\frac{1}{\left(2a+3b\right)^2}+\frac{1}{\left(2b+3a\right)^2}\right)\)

Nó đủ để ta có thể thấy rằng 

\(25ab\left[\left(2b+3a\right)^2+\left(2a+3b\right)^2\right]\le2\left(2a+3b\right)^2\left(2b+3a\right)^2\)

\(\Leftrightarrow59\left(a^2-b^2\right)^2+13\left(a^4+b^4-a^3b-ab^3\right)\ge0\)

BĐT cuối cùng đúng nên ta có ĐPCM

ok jjj

Đặt \(\frac{a}{b}=t\)do a>0, b>0 nên t>0

Khi đó BĐT \(\frac{2a^2+3b^2}{2a^3+3b^3}+\frac{2b^2}{3b^3}+\frac{2b^2+3a^2}{2b^3+3a^2}\le\frac{4}{a+b}\left(1\right)\)trở thành

\(\frac{2t^2+3}{2t^3+3}+\frac{2+3t^2}{3+3t^3}\le\frac{4}{t+1}\)

\(\Leftrightarrow\left(2t^2+3\right)\left(2+3t^2\right)\left(t+1\right)+\left(2+3t^2\right)\left(2t^2+1\right)\left(t+1\right)\le4\left(2t^3+3\right)\left(2+3t^2\right)\)

\(\Leftrightarrow\left(t+1\right)\left(12t^5+13t^3+13t^2+12\right)\le4\left(6t^6+13t^3+6\right)\)

\(\Leftrightarrow12\left(t^6-t^5-t+1\right)-13t^2\left(t^2-12t+1\right)\ge0\)

\(\Leftrightarrow12\left(t-1\right)^2\left[12\left(t^4+t^3+t^2+t+1\right)-13t^2\right]\ge0\)

\(\Leftrightarrow\left(t-1\right)^2\left[12\left(t^4+t^3+t^2+t+1\right)-13t^2\right]\ge0\left(2\right)\)

Ta có \(12\left(t^4+t^3+t^2+t+1\right)-13t^2=12t^4+12t\left(t-1\right)^2+23t^2+12>0\forall t>0\)

BĐT (2) đúng với mọi t>0

=> BĐT (1) đúng với mọi a,b>0

Dấu "=" xảy ra <=> t=1 <=> a=b

Áp dụng bđt Cauchy-schwarz dạng engel ta có:

1. \(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{\left(a+b+c\right)^2}{\left(a+2b\right)+\left(b+2c\right)+\left(c+2a\right)}=\frac{a+b+c}{3}\)

Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)

2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)

Dấu "=" \(\Leftrightarrow a=b=c\)