Chứng minh (x+2)^2+2/(x+2)>=3

Bài 5:

Áp dụng BĐT Cauchy cho $3$ số:

\((x+2)^2+\frac{2}{x+2}=(x+2)^2+\frac{1}{x+2}+\frac{1}{x+2}\geq 3\sqrt[4]{(x+2)^2.\frac{1}{x+2}.\frac{1}{x+2}}=3\)

Ta có đpcm

Dấu bằng xảy ra khi \((x+2)^2=\frac{1}{x+2}\Leftrightarrow x=-1\) (loại do $x>0$)

Do đó dấu bằng không xảy ra, hay \((x+2)^2+\frac{2}{x+2}>3\)

Bài 3)

Áp dụng BĐT Cauchy cho $5$ số ta có:

\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\geq 5\sqrt[5]{\frac{1}{b^{15}}}=\frac{5}{b^3}\)

Tương tự:

\(\frac{b^2}{c^5}+\frac{b^2}{c^5}+\frac{b^2}{c^5}+\frac{1}{b^3}+\frac{1}{b^3}\geq \frac{5}{c^3}\)

\(\frac{c^2}{d^5}+\frac{c^2}{d^5}+\frac{c^2}{d^5}+\frac{1}{c^3}+\frac{1}{c^3}\geq \frac{5}{d^3}\)

\(\frac{d^2}{a^5}+\frac{d^2}{a^5}+\frac{d^2}{a^5}+\frac{1}{d^3}+\frac{1}{d^3}\geq \frac{5}{a^3}\)

Cộng các BĐ vừa thu được:

\(\Rightarrow 3\left(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\right)+2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\right)\geq 5\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\right)\)

\(\Rightarrow \frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\geq \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c=d$