BĐT cần chứng minh tương đương với

\(\left(1-\frac{a^5-a^2}{a^5+b^2+c^2}\right)+\left(1-\frac{b^5-b^2}{b^5+c^2+a^2}\right)+\left(1-\frac{c^5-c^2}{c^5+a^2+b^2}\right)\le3\)

hay \(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Từ \(abc\ge1\) ta có:

\(\frac{1}{a^5+b^2+c^2}\le\frac{1}{\frac{a^5}{abc}+b^2+c^2}=\frac{1}{\frac{a^4}{bc}+b^2+c^2}\)

\(\le\frac{1}{\frac{2a^4}{b^2+c^2}+b^2+c^2}=\frac{b^2+c^2}{2a^4+\left(b^2+c^2\right)^2}\)

Do \(4u^2+v^2\ge4uv\Leftrightarrow4u^2+v^2\ge\frac{2}{3}\left(u+v\right)^2\)nên 

\(2a^4+\left(b^2+c^2\right)^2\ge\frac{2}{3}\left(a^2+b^2+c^2\right)^2\)

Suy ra \(\frac{1}{a^5+b^2+c^2}\le\frac{3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Tương tự ta có \(\frac{1}{b^5+c^2+a^2}\le\frac{3\left(c^2+a^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

và \(\frac{1}{c^5+a^2+b^2}\le\frac{3\left(a^2+b^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Cộng ba vế của các BĐT trên ta được

\(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Vậy \(\frac{a^5-a^2}{a^5+b^2+c^2}+\frac{b^5-b^2}{b^5+c^2+a^2}+\frac{c^5-c^2}{c^5+a^2+b^2}\ge0\)

(Dấu "="\(\Leftrightarrow a=b=c=1\))

Lời giải:

Do $ab+bc+ac=5$ nên:

\(a^2+5=a^2+ab+bc+ac=(a+b)(a+c)\)

\(b^2+5=b^2+ab+bc+ac=(b+c)(b+a)\)

\(c^2+5=c^2+ab+bc+ac=(c+a)(c+b)\)

Do đó:

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

\(=a\sqrt{(b+c)^2}+b\sqrt{(c+a)^2}+c\sqrt{(a+b)^2}=a(b+c)+b(c+a)+c(a+b)\)

\(=2(ab+bc+ac)=2.5=10\)

Vì vai trò của a,b,c như nhau,không mất tính tổng quát ta có:\(a\le b\le c\le1\Rightarrow\hept{\begin{cases}a-1\le0\\b-1\le0\\c-1\le0\end{cases}}\)

Áp dụng BĐT Cô-si ta có:

\(\frac{a^2}{a^2+b^5+c^5}\le\frac{a^2}{3\sqrt[3]{a^2b^5c^5}}=\frac{a^2}{3bc}\)

Tương tự:\(\frac{b^2}{b^2+a^5+c^5}\le\frac{b^2}{3ac};\frac{c^2}{c^2+a^5+b^5}\le\frac{c^2}{3ab}\)

Cộng vế với vế của 3 BĐT trên ta đươc:

\(\frac{a^2}{a^2+b^5+c^5}+\frac{b^2}{b^2+a^5+c^5}+\frac{c^2}{c^2+a^5+b^5}\le\frac{a^2}{3bc}+\frac{b^2}{3ac}+\frac{c^2}{3ab}=\frac{a^3+b^3+c^3}{3}\)

Xét \(a^3+b^3+c^3\le3\)

\(\Leftrightarrow\left(a^3-1\right)+\left(b^3-1\right)+\left(c^3-1\right)\le0\)

\(\Leftrightarrow\left(a-1\right)\left(a^2+a+1\right)+\left(b-1\right)\left(b^2+b+1\right)+\left(c-1\right)\left(c^2+c+1\right)\le0\) (đúng)

Từ đó suy ra:  

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

Dấu '='xảy ra khi\(\hept{\begin{cases}a=b=c\\abc=1\end{cases}\Leftrightarrow a=b=c=1}\)

\(a.S=2+2^2+2^3+...+2^{20}\\2S=2^2+2^3+...+2^{21}\\ 2S-S=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+2^3+...+2^{20}\right)\\ S=2^{21}-2\\ b,A=5+5^2+5^3+...+5^{96}\\ 5A=5^2+5^3+5^4+.......+5^{97}\\ 5A-A=\left(5^2+5^3+...+5^{97}\right)-\left(5+5^2+5^3+...+5^{96}\right)\\ 4A=5^{97}-5\\ A=\dfrac{5^{97}-5}{4}\)

\(S=2+2^2+2^3+...+2^{20}\)

\(\Rightarrow S=2\left(1+2^1+2^2+...+2^{19}\right)\)

\(\Rightarrow S=2.\dfrac{2^{19+1}-1}{2-1}=2\left(2^{20}-1\right)\)

\(B=5+5^2+5^3+...+5^{96}\)

\(\Rightarrow B=5\left(1+5^1+5^2+...+5^{95}\right)\)

\(\Rightarrow B=5.\dfrac{5^{95+1}-1}{5-1}=\dfrac{5\left(5^{96}-1\right)}{4}\)

đề bài yêu cầu là gì vậy

a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)

b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)

c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)

d)\(\dfrac{a}{b}=3\times\dfrac{2}{7}=\dfrac{6}{7}\)

e)\(\dfrac{a}{b}=\dfrac{7}{5}-\left(\dfrac{2}{5}\times\dfrac{1}{2}\right)=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{6}{5}\)

a 22/5

b 59/42

c 9/10

d 6/7

e 6/5

\(\frac{a^5-a^2}{a^5+b^2+c^2}+\frac{b^5-b^2}{b^5+c^2+a^2}+\frac{c^5-c^2}{c^5+a^2+b^2}\ge0\)

\(\Leftrightarrow1-\frac{a^2+b^2+c^2}{a^5+b^2+c^2}+1-\frac{a^2+b^2+c^2}{b^5+c^2+a^2}+1-\frac{a^2+b^2+c^2}{c^5+a^2+b^2}\ge0\)

\(\Leftrightarrow\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Áp dụng BĐT Cauchy-Schwarz ( chính là BĐT BCS) ta có:

\(\left(a^5+b^2+c^2\right)\left(\frac{1}{a}+b^2+c^2\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow\frac{1}{a^5+b^2+c^2}\le\frac{\frac{1}{a}+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}\).Tương tự:

\(\frac{1}{b^5+a^2+c^2}\le\frac{\frac{1}{b}+a^2+c^2}{\left(a^2+b^2+c^2\right)^2};\frac{1}{c^5+a^2+b^2}\le\frac{\frac{1}{c}+a^2+b^2}{\left(a^2+b^2+c^2\right)^2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=Σ\frac{1}{a^5+b^2+c^2}\le\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)^2}\)

Cần chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\) (Đúng)

Xảy ra khi \(a=b=c=1\)

-Lời giải được nhai lại từ Câu hỏi của LIVERPOOL - Toán lớp 9 - Học toán với OnlineMath

a,       S = 2 + 2² + 2³ + ...+ 2²⁰

     2S =      2² + 2³ +...+ 2²⁰ + 2²¹

2S - S =      2²¹ - 2

        S =      2²¹ - 2

b,         A =  5 + 5² + 5³ +...+ 5⁹⁶

         5A  =        5² + 5³ +...+ 5⁹⁶ + 5⁹⁷

     5A - A =        5⁹⁷ - 5

          4A =         5⁹⁷ - 5

          A   =         \(\dfrac{5^{97}-5}{4}\) 

\(\Leftrightarrow-a^{10}+a^{20}=0\)

=>a(a-1)(a+1)=0

hay \(a\in\left\{0;-1;1\right\}\)

`<=>(a^5. a^5)+[-(a^2. a^2)]^5=0`

`<=>a^10-a^20=0`

`<=>a^10(1-a^10)=0`

`<=>a^10=0` hoặc `1-a^10=0`

`<=>a=0` hoặc `a=1` hoặc `a=-1`