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a) \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)-\left(a+b+c\right)\left(ab+bc+ac\right)\)
\(=a^3+ab^2+ac^2+a^2b+b^3+c^2b+a^2c+b^2c+c^3-a^2b-abc-a^2c-ab^2-b^2c-abc-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc\left(đpcm\right)\)
b) Bạn chỉ cần nhân bung cả 2 vế ra là được á .
c) \(2\left(a+b+c\right)\left(\dfrac{b}{2}+\dfrac{c}{2}-\dfrac{a}{2}\right)\)
\(=2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)
\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)
1) a³ + b³ + c³ - 3abc
=(a + b)(a² - ab + b²) + c³ - 3abc
=(a + b)(a² - ab + b²) + c(a² - ab + b²) - 2abc - ca² - cb²
=(a + b + c)(a² - ab + b²) - (abc + b²c + bc² + ac² + abc + c²a) + c³ + ac² + bc²
=(a + b = c)(a² - ab + b²) - (a + b + c)(bc + ca) + c²(a + b + c)
=(a + b + c)(a² + b² + c² - ab - bc - ca)
2) \(\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)=\left(3a+5\right)\left(a-3\right)+2\left(7b-10\right)\left(1\right)\)
\(\Leftrightarrow3a^2+15a+2ab+10b-a-5-2ab+4b=3a^2+14a+15+14b-10\)
\(\Leftrightarrow3a^2+14a+14b-5=3a^2+14a+14b-5\)( đúng)
\(\Rightarrow\left(1\right)\) đúng (đpcm)
\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)
Ta chứng minh được \(ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
\(VT=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=VP\)
Vậy ta cần chứng minh \(2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge4a^2b^2c+4a^2bc^2+4ab^2c^2\)
\(\Leftrightarrow\sum_{cyc}\left(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b\right)\left(a-b\right)^2\ge0\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Câu a, b, c giống dạng nhau nên mình làm một câu a và câu d thôi nha, bạn tham khảo ^^
Giải:
a) \(a=\dfrac{b}{2}=\dfrac{c}{3}\)
Áp dụng tính chất của dãy tỉ sô bằng nhau:
\(a=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{a-b+c}{1-2+3}=\dfrac{10}{2}=5\)
\(\Rightarrow\left\{{}\begin{matrix}a=5.1=5\\b=2.5=10\\c=3.5=15\end{matrix}\right.\)
b) \(a:b:c=3:4:5\)
\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a^2}{9}=\dfrac{b^2}{16}=\dfrac{c^2}{25}\)
\(\Rightarrow\dfrac{2a^2}{18}=\dfrac{2b^2}{32}=\dfrac{3c^2}{75}\)
Áp dụng tính chất của dãy tỉ sô bằng nhau:
\(\Rightarrow\dfrac{2a^2}{18}=\dfrac{2b^2}{32}=\dfrac{3c^2}{75}=\dfrac{2a^2+2b^2-3c^2}{18+32-75}=\dfrac{-100}{-25}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=\dfrac{4.18}{2}=36\\b^2=\dfrac{4.32}{2}=64\\c^2=\dfrac{4.75}{3}=100\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\pm6\\b=\pm8\\c=\pm10\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)
Theo hệ quả của BĐT AM-GM ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
a: a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3ab(a+b)-3bac
=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
b: Đề sai rồi bạn
c: 2(a+b+c)*(b/2+c/2-a/2)
=(a+b+c)(b+c-a)
=(b+c)^2-a^2
=c^2+2bc+c^2-a^2