1/a+a/4+3/4.a 

2 cái đầu dùng cosi,,cái 3 dùng a>=2

dấu = khi a=2

ta có A=a+1/a=1+(1/a)

có:a>=2 =>min=3/2 tại a=2

bdt co si =>minA=2

A=\(x^2-\frac{1}{3}x+1=x^2-2.\frac{1}{6}.x+\frac{1}{36}-\frac{1}{36}+1\)

\(=\left(x+\frac{1}{6}\right)^2+\frac{35}{36}\)

Do \(\left(x+\frac{1}{6}\right)^2\ge0\)nên \(\left(x+\frac{1}{6}\right)^2+\frac{35}{36}>0\)và GTNN của A là  \(\frac{35}{36}\)

hình như cái khúc (x+1/2)^2 phải là (x-1/2)^2 chứ bạn mk k hỉu rõ bạn giải thích giùm mk nhé

à.mk nhìn nhầm

\(\left(x-\frac{1}{6}\right)^2\) mới  đúng.

\(A=x-2\sqrt{x}\left(\sqrt{y}+1\right)+\left(\sqrt{y}+1\right)^2+\left(3y+1-\left(\sqrt{y}+1\right)^2\right)\)

 \(=\left(\sqrt{x}-\sqrt{y}-1\right)^2+2\left(y-\sqrt{y}+\frac{1}{4}\right)-\frac{1}{2}\)

\(=\left(\sqrt{x}-\sqrt{y}-1\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)

Amin= -1/2  khi  y=1/4; x=9/4

\(M=a+b+\frac{1}{a}+\frac{1}{b}\ge a+b+\frac{4}{a+b}=a+b+\frac{1}{a+b}+\frac{3}{a+b}\)

\(\Rightarrow M\ge2\sqrt{\frac{a+b}{a+b}}+3=5\)

\(\Rightarrow M_{min}=5\) khi \(a=b=\frac{1}{2}\)

\(A=a^2+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\ge\dfrac{3.2}{4}+1=\dfrac{5}{2}\)

Vậy GTNN là \(A=\dfrac{5}{2}\) dấu = xảy ra khi \(a^2=2\)

Ta có: \(A=a^2+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\dfrac{a^2}{4}+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\)

Do \(a^2\ge2\) => \(\dfrac{3a^2}{4}\ge\dfrac{3}{4}.2=\dfrac{3}{2}\) (*)

Áp dụng BĐT cô-si :

\(\dfrac{a^2}{4}+\dfrac{1}{a^2}\ge2\sqrt{\dfrac{a^2}{4}.\dfrac{1}{a^2}}=2.\dfrac{1}{2}=1\) (**)

Từ (*) và (**) suy ra :

\(\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\ge\dfrac{3}{2}+1=\dfrac{5}{2}\)

<=> \(A\ge\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(a^2=2\) <=> \(a=\pm\sqrt{2}\)

Vậy GTNN của \(A=a^2+\dfrac{1}{a^2}\) là \(\dfrac{5}{2}\) khi \(a=\pm\sqrt{2}\)

\(P=a^3+b^3+c^3+a^2\left(b+c\right)+b^2\left(a+c\right)+c^2\left(a+b\right)\)

\(=a^3+b^3+c^3+a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\)

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

Đạt được khi \(a=b=c=\frac{1}{3}\)

\(a=\dfrac{a+1}{a-2020}\)

\(=\dfrac{a-2020}{a-2020}+\dfrac{2021}{a-2020}\)

\(=1+\dfrac{2021}{a-2020}\) Vì a>2020

⇒\(1+\dfrac{2021}{a-2020}\text{≥}2\)

Min a=2 ⇔\(\dfrac{2021}{a-2020}=1\)

                ⇔\(a-2020=2021\)

                ⇔\(a=4041\)