a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)

Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)

Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)

* Các BĐT phụ bạn tự CM nha! Chúc bạn học tốt

Camon bạn!!! Nhưng bạn đọc sai đề r !! ^.^

Câu 9.

a) Ta có: \(\left(a-1\right)^2\ge0\)(điều hiển nhiên)

\(\Leftrightarrow a^2-2a+1\ge0\)

\(\Leftrightarrow a^2+2a+1\ge4a\)

\(\Leftrightarrow\left(a+1\right)^2\ge4a\left(đpcm\right)\)

b) Áp dụng BĐT Cauchy cho 2 số không âm:

\(a+1\ge2\sqrt{a}\)

\(b+1\ge2\sqrt{b}\)

\(c+1\ge2\sqrt{c}\)

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(Vì abc = 1)

Câu 10. 

a) Ta có: \(-\left(a-b\right)^2\le0\)(điều hiển nhiên)

\(\Leftrightarrow-a^2+2ab-b^2\le0\)

\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

Có: \(2ab\le a^2+b^2;2bc\le b^2+c^2;2ac\le a^2+c^2\)(BĐT Cauchy)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3\left(a^2+b^2+c^2\right)\)

Vậy ​​\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

cho a b c 0 và a+b+c=3 CMR a/1+b^2 +b/1+c^2 +c/1+a^2 >=3/2

\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)

\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)

\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)

Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)

\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)

Rồi tương tự các kiểu:v

Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)

\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))

Không phải dùng tới Cauchy-Schwarz:D

nhầm làm lại nha ^^

(a+b+c)^2=a^2+b^2+c^2

=>a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

=>2(ab+bc+ac)=0

=>ab+bc+ac=0

=>(ab+bc+ac)/abc=0

=>ab/abc+bc/abc+ac/abc=0

=>1/c+1/a+1/b=0

=> 1/a+1/b=-1/c

=> (1/a+1/b)^3=(-1/c)^3

=> 1/a^3+1/b^3+3/ab(1/a+1/b)=-1/c^3

=> 1/a^3+1/b^3+1/c^3+3/ab.(-1/c)=0

=> 1/a^3+1/b^3+1/c^3-3/abc=0

=> 1/a^3+1/b^3+1/c^3=3/abc (đpcm)

(a+b+c)^2=a^2+b^2+c^2

a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

2(ab+bc+ac)=0

ab+bc+ac=0

(ab+bc+ac)/abc=0

ab/abc+bc/abc+ac/abc=0

1/c+1/a+1/b=0

=> 1/a+1/b=-1/c

=> (1/a+1/b)^3=(-1/c)^3

=> 1/a^3+1/b^3+3.(1/a.)(1/b).(1/a+1/b)=-1/c^3

=> 1/a^3+1/b^3+1/c^3.3ab.(-1/c)=0

=> 1/a^3+1/b^3+1/c^3=3/abc

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

\(\ge2\sqrt{a^2.\frac{1}{4}}+2\sqrt{b^2.\frac{1}{4}}-\frac{1}{2}\) (bdt cosi)

\(=a+b-\frac{1}{2}=1-\frac{1}{2}=\frac{1}{2}\) (vi a+b=1)

dau = xay ra <=> a=b=1/2

chuc ban hoc tot

mik phai di ngu nen lam hoi tat mong bn thong cam

phan b bn lam tuong tu nha

1/ Ta có:

\(\left(a-b\right)^2\ge0,\) mọi a, b

<=> \(a^2-2ab+b^2\ge0\)

<=> \(2a^2+2b^2\ge a^2+2ab+b^2\)

<=> \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

Dấu bằng xảy ra <=>  a - b = 0 <=> a  = b.

2/ Dựa vào câu 1. 

\(a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(\frac{1}{2}\right)^2}{2}=\frac{1}{8}\).

1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự :  \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)

\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)

Từ   \(a=b+c\)   \(\Rightarrow\)  \(a-b-c=0\)    

Ta có:

\(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=1\)

\(\Rightarrow\)  \(\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a}{abc}-\frac{b}{abc}-\frac{c}{abc}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a-b-c}{abc}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c-c}{abc}\right)=1\)