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Ta có a+b+c=0 => \(a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3ab\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)
\(a^6+b^6+c^6=\left(a^3\right)^2+\left(b^3\right)^2+\left(c^3\right)^2=\left(a^3+b^3+c^3\right)^2-2\left(a^3b^3+b^3c^3+c^3a^3\right)\)
\(ab+bc+ca=0\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Do đó: \(a^6+b^6+c^6=\left(3abc\right)^2-2\cdot3a^2b^2c^2=3a^2b^2c^2\)
Vậy \(\frac{a^6+b^6+c^6}{a^3+b^3+c^3}=\frac{3a^2b^2c^2}{3abc}=abc\left(đpcm\right)\)
Áp dụng BĐT cosi ta có
\(\frac{a^6}{b^3}+\frac{b^6}{c^3}+1\ge3\sqrt[3]{\frac{a^6.b^3}{c^3}}=\frac{3a^2b}{c}\)
\(\frac{b^6}{c^3}+\frac{c^6}{a^3}+1\ge\frac{3b^2c}{a}\)
\(\frac{c^6}{a^3}+\frac{a^6}{b^3}+1\ge\frac{3c^2a}{b}\)
Cộng 3 bĐt trên
=> \(2.VT+3\ge3\left(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\right)=9\)
=> \(VT\ge3\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=1
\(1,\hept{\begin{cases}10x^2+5y^2-2xy-38x-6y+41=0\left(1\right)\\3x^2-2y^2+5xy-17x-6y+20=0\left(2\right)\end{cases}}\)
Giải (1) : \(10x^2+5y^2-2xy-38x-6y+41=0\)
\(\Leftrightarrow10x^2-2x\left(y+19\right)+5y^2-6y+41=0\)
Coi pt trên là pt bậc 2 ẩn x
Có \(\Delta'=\left(y+19\right)^2-50y^2+60y-410\)
\(=-49y^2+98y-49\)
\(=-49\left(y-1\right)^2\)
pt có nghiệm \(\Leftrightarrow\Delta'\ge0\)
\(\Leftrightarrow-49\left(y-1\right)^2\ge0\)
\(\Leftrightarrow y=1\)
Thế vào pt (2) được x = 2
\(2,\)Đặt\(\left(a\sqrt{a};b\sqrt{b};c\sqrt{c}\right)\rightarrow\left(x;y;z\right)\left(x,y,z>0\right)\)
\(\Rightarrow xy+yz+zx=1\)
Khi đó \(P=\frac{x^4}{x^2+y^2}+\frac{y^4}{y^2+z^2}+\frac{z^4}{x^2+z^2}\)
Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(x;y;z>0\right)\left(Cauchy-engel-type_3\right)\)được
\(P\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{2}\)
Áp dụng bđt x2 + y2 + z2 > xy + yz + zx (tự chứng minh) ta được
\(P\ge\frac{x^2+y^2+z^2}{2}\ge\frac{xy+yz+zx}{2}=\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}xy+yz+zx=1\\x=y=z\end{cases}}\)
\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\sqrt{a^3}=\sqrt{b^3}=\sqrt{c^3}=\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow a^3=b^3=c^3=\frac{1}{3}\)
\(\Leftrightarrow a=b=c=\frac{1}{\sqrt[3]{3}}\)
Vậy \(P_{min}=\frac{1}{2}\Leftrightarrow a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(P=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)(BĐT Svarxơ)\(\ge\frac{\frac{1}{9}\left(a+b+c\right)^4}{ab+bc+ca}\)(BĐT Bunhiacoxki)
Có: \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow ab+bc+ca\le3\)
\(\Rightarrow P\ge\frac{\frac{1}{9}\left(a+b+c\right)^4}{3}\)\(=\frac{1}{27}\left(a+b+c\right)^4\)
Dễ thấy \(P\ge3\)
Cần C/m \(\left(a+b+c\right)^4\ge81\)
\(\Rightarrow a+b+c\ge3\)
mà\(ab+bc+ca\le3\) kết hợp với gt nên ta có điều đó LĐ.
Vậy Pmin=3\(\Leftrightarrow a=b=c=1\)
Ta luôn có: \(ab+ac+bc\le\frac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow a+b+c+\frac{\left(a+b+c\right)^2}{3}\ge6\)
\(\Rightarrow\left(a+b+c\right)^2+3\left(a+b+c\right)-18\ge0\)
\(\Rightarrow\left(a+b+c-3\right)\left(a+b+c+6\right)\ge0\)
\(\Rightarrow a+b+c-3\ge0\) (do \(a+b+c+6>0\))
\(\Rightarrow a+b+c\ge3\)
\(P=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+ac+bc}\ge\frac{\left(\frac{\left(a+b+c\right)^2}{3}\right)^2}{\frac{\left(a+b+c\right)^2}{3}}=\frac{\left(a+b+c\right)^2}{3}\ge3\)
\(\Rightarrow P_{min}=3\) khi \(a=b=c=1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(F=\frac{a^6}{b^3+c^3}+\frac{b^6}{c^3+a^3}+\frac{c^6}{a^3+b^3}\)
\(\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)}=\frac{a^3+b^3+c^3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(a^3+\frac{1}{27}+\frac{1}{27}\ge3\sqrt[3]{a^3\cdot\frac{1}{27}\cdot\frac{1}{27}}=3\cdot\frac{a}{9}=\frac{a}{3}\)
Tương tự ta cũng có: \(b^3+\frac{1}{27}+\frac{1}{27}\ge\frac{b}{3};c^3+\frac{1}{27}+\frac{1}{27}\ge\frac{c}{3}\)
\(\Rightarrow a^3+b^3+c^3+\frac{2}{9}\ge\frac{a+b+c}{3}=\frac{1}{3}\Rightarrow a^3+b^3+c^3\ge\frac{1}{9}\)
\(\Rightarrow F\ge\frac{a^3+b^3+c^3}{2}\ge\frac{\frac{1}{9}}{2}=\frac{1}{18}\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
lồn to