1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)

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

\(\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2ab}\ge\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}=6\)

\("="\Leftrightarrow a=b=\frac{1}{2}\)

2) \(S=a+\frac{1}{a}=\frac{15a}{16}+\left(\frac{a}{16}+\frac{1}{a}\right)\)

Áp dụng BĐT AM-GM ta có:

\(S\ge\frac{15a}{16}+2.\sqrt{\frac{a}{16}.\frac{1}{a}}=\frac{15.4}{16}+2.\sqrt{\frac{1}{16}}=\frac{15}{4}+2.\frac{1}{4}=\frac{15}{4}+\frac{1}{2}=\frac{15}{4}+\frac{2}{4}=\frac{17}{4}\)

\(S=\frac{17}{4}\Leftrightarrow a=4\)

Vậy \(S_{min}=\frac{17}{4}\Leftrightarrow a=4\)

kudo shinichi sao cách làm giống của thầy Hồng Trí Quang vậy bạn?

\(S=a+\frac{1}{a}=\frac{15}{16}a+\left(\frac{a}{16}+\frac{1}{a}\right)\ge\frac{15}{16}a+2\sqrt{\frac{1.a}{16.a}}=\frac{15}{16}a+2.\frac{1}{4}\)

\(=\frac{15}{16}.4+\frac{1}{2}=\frac{17}{4}\Leftrightarrow a=4\)

Dấu "=" xảy ra khi a = 4

Vậy \(S_{min}=\frac{17}{4}\Leftrightarrow a=4\)

\(a\ge2b\Rightarrow\dfrac{a}{b}\ge2\)

\(A=\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a}{4b}+\dfrac{b}{a}+\dfrac{3}{4}.\dfrac{a}{b}\ge2\sqrt{\dfrac{ab}{4ab}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)

\(A_{min}=\dfrac{5}{2}\) khi \(a=2b\)

B1 

Ta có

\(A=\frac{a^2}{24}+\frac{9}{a}+\frac{9}{a}+\frac{23a^2}{24}\ge3\sqrt[3]{\frac{a^2}{24}.\frac{9}{a}.\frac{9}{a}+\frac{23a^2}{24}}\ge\frac{9}{2}+\frac{23.36}{24}\ge39\)

Dấu "=" xảy ra <=> a=6

Vậy Min A = 39 <=> a=6

 \(A=a^2+\frac{18}{a}=a^2+\frac{216}{a}+\frac{216}{a}-\frac{414}{a}\ge3\sqrt[3]{a^2.\frac{216}{a}.\frac{216}{a}}-69=39\)

Đẳng thức xảy ra khi a = 6

B3: Áp dụng bđt AM-GM

\(A=\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}+\frac{3\left(a+b\right)}{4\sqrt{ab}}\ge2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}+\frac{3\left(a+b\right)}{4\left(\frac{a+b}{2}\right)}\)

\(=1+\frac{3\left(a+b\right)}{2\left(a+b\right)}=1+\frac{3}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(a=b>0\)

-Đề thiếu?

\(P=\left(a^2+\dfrac{1}{16a^2}\right)+\left(b^2+\dfrac{1}{16b^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge2\sqrt{\dfrac{a^2}{16a^2}}+2\sqrt{\dfrac{b^2}{16b^2}}+\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)

\(P\ge1+\dfrac{15}{32}.\left(\dfrac{4}{a+b}\right)^2\ge1+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=\dfrac{17}{2}\)

\(P_{min}=\dfrac{17}{2}\) khi \(a=b=\dfrac{1}{2}\)