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\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
\(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\)
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
mà \(a,b,c\)dương nên \(x=y=z\Rightarrow a=b=c\).
\(A=\left(2+\frac{a}{b}\right)\left(2+\frac{b}{c}\right)\left(2+\frac{c}{a}\right)=3^3=27\).
\(3a^2\)\(b^2\)\(c^2\)
\(=>ab+bc+ca=0\)
\(=>ab^2\)\(+bc^2\)\(+ca^2\)\(=0\)
\(TH1:ab+bc+ca=0\)
\(ab+bc=-ca\)
\(=>a+c=-\frac{ac}{b}\)
\(=>a+b=-\frac{ab}{c}\)
\(b+c=-\frac{bc}{a}\)
\(Thay\)\(A\)
\(=>A=-3\)
\(\left(ab-bc\right)^2\)\(+\left(bc-ca\right)^2\)\(+\left(ca-ab\right)^2\)\(=0\)
\(=>ab-bc=0\)
\(bc-ca=0\)
\(ca-ab=0\)
\(=>ab=bc=ca\)
\(=>a=b=c\)
\(Thay\)\(A\)
\(=>A=-24\)
\(=>A=\left(-3;-24\right)\)
Em làm sai mong anh thông cảm cho ạ
Ta co:
\(\frac{x^4}{a}+\frac{y^4}{b}\ge\frac{\left(x^2+y^2\right)^2}{a+b}=\frac{1}{a+b}\)
Dau '=' xay ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\)
Ta lai co:
\(\frac{x^6}{a^3}+\frac{y^6}{b^3}=\left(\frac{x^2}{a}\right)^3+\left(\frac{y^2}{b}\right)^3=2\left(\frac{x^2}{a}\right)^3\)
Ma \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow x^2=\frac{a}{a+b}\)
\(\Leftrightarrow\frac{x^2}{a}=\frac{1}{a+b}\)
\(\Leftrightarrow\left(\frac{x^2}{a}\right)^3=\frac{1}{\left(a+b\right)^3}\)
\(\Rightarrow\frac{x^6}{a^3}+\frac{y^6}{b^3}=\frac{2}{\left(a+b\right)^3}\)
Bài 2:
A = (a+b)(1/a+1/b)
Có: \(a+b\ge2\sqrt{ab}\)
\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\)
=> \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4\)
=> ĐPCM
1.b)
Pt (1) : 4(n + 1) + 3n - 6 < 19
<=> 4n + 4 + 3n - 6 < 19
<=> 7n - 2 < 19
<=> 7n - 2 - 19 < 0
<=> 7n - 21 < 0
<=> n < 3
Pt (2) : (n - 3)^2 - (n + 4)(n - 4) ≤ 43
<=> n^2 - 6n + 9 - n^2 + 16 ≤ 43
<=> -6n + 25 ≤ 43
<=> -6n ≤ 18
<=> n ≥ -3
Vì n < 3 và n ≥ -3 => -3 ≤ n ≤ 3.
Vậy S = {x ∈ R ; -3 ≤ n ≤ 3}