Ta có: \(ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow\left(a+b\right)^2\ge4\Rightarrow a+b\ge2\)

Và \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\ge2\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^3}{b+1}=a^3-\frac{a^3b}{b+1}\ge a^3-\frac{a^3b}{2\sqrt{b}}=a^3-\frac{a^3\sqrt{b}}{2}\)

Tương tự cho ta cũng có:\(\frac{b^3}{a+1}\ge b^3-\frac{b^3\sqrt{a}}{2}\)

\(\Rightarrow Q\ge a^3+b^3-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\left(1\right)\)

TIếp tục xài AM-GM: \(\sqrt{b}\le\frac{b+1}{2}\Rightarrow a^3\sqrt{b}=\frac{a^3b+a^3}{2}\)

\(\Rightarrow\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\le\frac{\frac{a^3b+a^3}{2}+\frac{ab^3+b^3}{2}}{2}=\frac{a^3b+ab^3+a^3+b^3}{4}\)

\(\Rightarrow2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\frac{a^3b+ab^3+a^3+b^3}{4}\)

Cần chứng minh \(2-\frac{a^3b+ab^3+a^3+b^3}{4}\ge1\)\(\Leftrightarrow\frac{a^3b+ab^3+a^3+b^3}{4}\ge1\)

\(\Leftrightarrow a^3b+ab^3+a^3+b^3\ge4\Leftrightarrow a^3b+ab^3\ge2\) vì \(a^3+b^3\ge2\)

\(\Leftrightarrow\left(ab\right)^2\left(a+b\right)\ge2\) đúng vì ab=1 và \(a+b\ge2\)

\(\Rightarrow Q_{Min}=2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-1=1\)

Khi a=b=1

So easy =))

Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:

\(F=\frac{4}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\)

\(\ge\frac{\left(1+2\right)^2}{2ab+a^2+b^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)

\(\ge\frac{\left(1+2\right)^2}{\left(a+b\right)^2}+\frac{\frac{\frac{\left(\left(a+b\right)^2\right)^2}{2}}{2}}{2}\)

\(=\frac{9}{1}+\frac{\frac{\frac{1}{2}}{2}}{2}=9+\frac{1}{8}=\frac{73}{8}\)

Xảy ra khi \(a=b=\frac{1}{2}\)

\(A=\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}-2+5}{\sqrt{x}-2}\)

\(A=1+\frac{5}{\sqrt{x}-2}\)

Để A nguyên\(\Leftrightarrow1+\frac{5}{\sqrt{x}-2}\)

mà 1 nguyên \(\Rightarrow\frac{5}{\sqrt{x}-2}\)nguyên

                           \(\sqrt{x}-2\in\text{Ư}\left(5\right)=5;-5;1;-1\)

Lập bảng là xong nhé

Mk cũng giống như các bn