\(\dfrac{4}{a+b}-\dfrac{2a^2+3b^2}{2a^3+3b^3}-\dfrac{2b^2+3a^2}{2b^3+3a^3}=\dfrac{\left(a-b\right)^2.\left(12b^4+12ab^3-a^2b^2+12a^3b+12a^4\right)}{\left(a+b\right)\left(2a^3+3b^3\right)\left(2b^3+3a^3\right)}\ge0\)

PS: Còn cách dùng holder nữa mà lười quá

holder Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath

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

a: \(\frac{x}{y}=\frac13\)

=>y=3x

\(\frac{14x+5y}{3x-11y}=\frac{11x+5\cdot3x}{3x-11\cdot3x}=\frac{26x}{3x-33x}=\frac{26x}{-30x}=\frac{-26}{30}=-\frac{13}{15}\)

b: \(\frac{a}{b}=\frac12\)

=>b=2a

\(\frac{11a^4-3ab^3+15a^3b+7b^4}{3a^2b^2+ab^3-6a^3b-2b^4}\)

=\(\frac{11a^4-3a\cdot\left(2a\right)^3+15a^3\cdot2a+7\left(2a\right)^4}{3a^2\cdot\left(2a\right)^2+a\cdot\left(2a\right)^3-6a^3\cdot2a-2\cdot\left(2a\right)^4}\)

\(=\frac{11a^4-3a\cdot8a^3+30a^4+7\cdot16a^4}{3a^2\cdot4a^2+a\cdot8a^3-6a^3\cdot2a-2\cdot16a^4}\)

\(=\frac{11a^4-24a^4+30a^4+112a^4}{12a^4+8a^4-12a^4-32a^4}=\frac{11-24+30+112}{12+8-12-32}=\frac{129}{-24}=\frac{-43}{8}\)

\(\Leftrightarrow\frac{\left(2a^2+3b^2\right)\left(a+b\right)}{2a^3+3b^3}+\frac{\left(2b^2+3a^2\right)\left(a+b\right)}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^3+3b^3+2a^2b+3ab^2}{2a^3+3b^3}+\frac{2b^3+3a^3+2ab^2+3ab^2}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^2b+3ab^2}{2a^3+3b^3}+\frac{2ab^2+3ab^2}{2b^3+3a^3}\le2\)

\(\Leftrightarrow\frac{2\left(\frac{a}{b}\right)^2+3\left(\frac{a}{b}\right)}{2\left(\frac{a}{b}\right)^3+3}+\frac{2\left(\frac{a}{b}\right)+3\left(\frac{a}{b}\right)^2}{3\left(\frac{a}{b}\right)^3+2}\le2\)

Đặt \(\frac{a}{b}=x>0\Rightarrow\frac{2x^2+3x}{2x^3+3}+\frac{3x^2+2x}{3x^3+2}\le2\)

\(\Leftrightarrow\left(x-1\right)^2\left(12x^4+12x^3-x^2+12x+12\right)\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=1\) hay \(a=b\)

Hơi trâu bò :D

A = 0

A=0