Áp dụng BĐT cô si với hai số không âm, Ta có: 

\(\left(a+b+c\right)^2=1\ge4a\left(b+c\right)\)

\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\)

Mà \(\left(b+c\right)^2\ge4bc\forall b,c\ge0\)

\(\Rightarrow b+c\ge16abc\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}a+b+c=1\\b=c\\a=b+c\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)

Áp dụng BĐT Cô si với 2 số dương ta có: 

\(\frac{a}{b}+\frac{b}{a}\ge2,\frac{b}{c}+\frac{c}{b}\ge2,\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)(đúng) 

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(do a+b+c=1)

theo đề ta có x=50 thuộc 90=70

-6869^=67

-78Y^7

TK: \(a^2+b^2+c^2+abc=4\Leftrightarrow a^2+abc+b^2+c^2-4=0\)

Coi đây là PT bậc 2 ẩn a

\(\Leftrightarrow\Delta=b^2c^2-4b^2-4c^2+16=\left(4-b^2\right)\left(4-c^2\right)\\ \Leftrightarrow a=\dfrac{-bc+\sqrt{\left(4-b^2\right)\left(4-c^2\right)}}{2}\le\dfrac{-bc+\dfrac{4-b^2+4-c^2}{2}}{2}=\dfrac{8-\left(b+c\right)^2}{4}\\ \Leftrightarrow a+b+c\le\dfrac{8-\left(b+c\right)^2}{4}+b+c=\dfrac{8-\left(b+c\right)^2+4\left(b+c\right)}{4}=\dfrac{12-\left(b+c-2\right)^2}{4}\le3\)

đặt \(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(=>A^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(=>A^2\le\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+a}\right)^2\right].3\)

\(=>A^2\le\left[2\left(a+b+c\right)\right]3=2.3=6\)

\(=>A\le\sqrt{6}\left(dpcm\right)\)

dấu"=" xảy ra<=>a=b=c=1/3

Ta có:\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2=\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\)

  \(\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)=3.2=6\)

\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)

Dấu "=" xảy ra <=> a=b=c=1/3

Gọi \(d=gcd\left(a;b\right)\) khi đó \(a=dm;b=dn\) với \(\left(m;n\right)=1\)

Ta có:

\(c+\frac{1}{b}=a+\frac{b}{a}\Leftrightarrow c=\frac{b}{a}+a-\frac{1}{b}=\frac{dn}{dm}+dm-\frac{1}{dn}\)

\(=\frac{n}{m}+dm-\frac{1}{dn}=\frac{dn^2+d^2m^2n-m}{dmn}\)

Khi đó \(dn^2+d^2m^2n-m⋮dmn\Rightarrow m⋮n\) mà \(\left(m;n\right)=1\Rightarrow n=1\Rightarrow m=d\)

Khi đó \(ab=dm\cdot dn=d^3\) là lập phương số nguyên dương

Với a > 0, b > 0, c > 0, d > 0 ta có:

a < b ⇒ ac < bc (1)

c < d ⇒ bc < bd (2)

Từ (1) và (2) suy ra: ac < bd.

c1:áp dụng bđt AM-GM:

\(a+b\ge2\sqrt{ab}\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2=1008^2\)

=> đáp án A

c2: tương tự c1 . đáp án b

3.

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ab}}=2\)

Đáp án A

4.

\(a^2-a+1=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) ;\(\forall a\)

Đáp án A