a, \(A=\left(a-b\right)+\left(a+b-c\right)-\left(a-b-c\right)\)

\(=a-b+a+b-c-a+b+c=a+b\)

b, \(B=\left(a-b\right)-\left(b-c\right)+\left(c-a\right)-\left(a-b-c\right)\)

\(=a-b-b+c+c-a-a+b+c=-a-b+3c\)

c, \(C=\left(-a+b+c\right)-\left(a-b+c\right)-\left(-a+b-c\right)\)

\(=-a+b+c-a+b-c+a-b+c=-a+b+c\)

1)   \(\left(a+b\right)-\left(-a+b-c\right)+\left(c-a-b\right)\)

\(=a+b+a-b+c+c-a-b\)

\(=a-b+2c \left(đpcm\right)\)

2)  \(a\left(b-c\right)-a\left(b+d\right)\)

\(=ab-ac-ab-ad\)

\(=-ac-ad\)

\(=-a\left(c+d\right) \left(đpcm\right)\)

Bài 1:

Theo bất đẳng thức Cauchy, ta có:

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

Chứng minh tương tự:

\(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) (2)

\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\) (3)

Từ (1), (2) và (3) suy ra:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{b+c}{4}+\dfrac{c+a}{4}+\dfrac{a+b}{4}\ge a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge a+b+c\)

Bài 2:

Theo bđt Cauchy ta có:

\(1+\dfrac{1}{a}=\dfrac{a+1}{a}=\dfrac{2a+b+c}{a}\ge\dfrac{2a+2\sqrt{bc}}{a}\ge\dfrac{2\left(a+\sqrt{bc}\right)}{a}\ge\dfrac{4\sqrt{a\sqrt{bc}}}{a}\)

\(\Rightarrow1+\dfrac{1}{a}\ge4\sqrt[4]{\dfrac{bc}{a^2}}\)

Chứng minh tương tự:

\(1+\dfrac{1}{b}\ge4\sqrt[4]{\dfrac{ca}{b^2}}\)

\(1+\dfrac{1}{c}\ge4\sqrt[4]{\dfrac{ab}{c^2}}\)

Nhân vế theo vế 3 bđt trên ta được:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge4^3\sqrt[4]{\dfrac{\left(abc\right)^2}{a^2b^2c^2}}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\left(dpcm\right)\)

bạn có thể tham khảo các câu hỏi tương tự

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

\(\cdot\frac{a}{b}=1\Rightarrow a=b\left(1\right)\)

\(\cdot\frac{b}{c}=1\Rightarrow b=c\left(2\right)\)

\(\cdot\frac{c}{a}=1\Rightarrow c=a\left(3\right)\)

\(\text{Từ (1);(2) và (3) suy ra }a=b=c\left(\text{ĐPCM}\right)\)

Lời giải:
Có:
$(a^2+1)(b^2+1)(c^2+1)=(a^2+ab+bc+ac)(b^2+ab+bc+ac)(c^2+ab+bc+ac)$

$=(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)=[(a+b)(b+c)(c+a)]^2$

Và:

$(a+b+c-abc)^2=[(a+b+c)(ab+bc+ac)-abc]^2$

$=[ab(a+b)+bc(b+c)+ca(c+a)+2abc]^2$

$=[ab(a+b+c)+bc(b+c+a)+ca(c+a)]^2$

$=[(a+b+c)(ab+bc)+ca(c+a)]^2=[b(a+b+c)(a+c)+ac(c+a)]^2$

$=[(c+a)(ab+b^2+bc+ac)]^2=[(c+a)(b+a)(b+c)]^2$
Do đó: $P=\frac{[(a+b)(b+c)(c+a)]^2}{[(a+b)(b+c)(c+a)]^2}=1$

Khi a=-4/5

= > A=-4/5.1/2+(-4/5).1/3+(-4/5).1/4

A=-4/5.(1/2+1/3-1/4)

A=-4/5.7/12

A=-7/15

Các bài còn lai tương tự

Áp dụng BĐT Cauchy - Schawrz dạng Engel, ta có: 

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

Vậy..

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

Ta có: \(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)(2)

Từ (1), (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\left(=\frac{a^2}{c^2}\right)\)

đang định lướt qua à,dừng lại giúp đê

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