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\(P=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)(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 \(a=b=c=1\)
bài này easy thôi:
Áp dụng BĐT schwarz ta có:
\(VT=\frac{a^4}{a\left(a^2+ab+b^2\right)}+\frac{b^4}{b\left(b^2+bc+c^2\right)}+\frac{c^4}{c\left(c^2+ac+a^2\right)}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ac+a^2\right)}.\)
Mặt khác \(a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ac+a^2\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right).\)
nên ta có:\(VT\ge\frac{a^2+b^2+c^2}{a+b+c}=a^2+b^2+c^2.\)
Mà ta có BĐT cơ bản là:\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2.\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge1\Leftrightarrow a^2+b^2+c^2\ge\frac{1}{3}.\)
Do đó:\(VT\ge a^2+b^2+c^2\ge\frac{1}{3}.\)
Vậy Min là \(\frac{1}{3}.\)Dấu = xảy ra khi \(a=b=c=\frac{1}{3}.\)
Ta có: \(P=\Sigma\frac{\left(\frac{1}{c^2}\right)}{\left(\frac{1}{a}+\frac{1}{b}\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\ge\frac{\left(\frac{9}{a+b+c}\right)}{2}=\frac{3}{2}\)
Đẳng thức xảy ra khi a =b =c = 1.
True?
Ta có :
\(P=\frac{ab}{c^2\left(a+b\right)}+\frac{ac}{b^2\left(a+c\right)}+\frac{bc}{a^2\left(b+c\right)}\)
\(\Rightarrow P=\frac{\left(\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}}+\frac{\left(\frac{1}{b}\right)^2}{\frac{1}{c}+\frac{1}{a}}+\frac{\left(\frac{1}{a}\right)^2}{\frac{1}{c}+\frac{1}{b}}\)
\(\Rightarrow P\ge\frac{\left(\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}}\)
\(\Rightarrow P\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(\Rightarrow P\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\ge\frac{1}{2}.\frac{9}{a+b+c}\)
\(\Rightarrow P\ge\frac{3}{2}\)
Dấu = xảy ra khi a=b=c=1
BĐt phụ : \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
c/m :\(3a^2-3ab+3b^2\ge a^2+ab+b^2\)
↔\(2a^2-4ab+2b^2\ge0\)
↔\(2\left(a-b\right)^2\ge0\)(luôn đúng)
Giải ;
ta có:\(\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)
→\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}\)(1)
mà \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\Leftrightarrow\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\ge\frac{1}{3}\left(a+b\right)\)
↔\(\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{1}{3}\left(a+b\right)\)
tương tự ta có:\(\frac{b^3+c^3}{b^2+bc+c^2}\ge\frac{1}{3}\left(b+c\right)\);\(\frac{c^3+a^3}{c^2+ca+a^2}\ge\frac{1}{3}\left(a+c\right)\)
cộng vế vs vế ta có:
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}+\frac{a^3}{c^2+ac+a^2}\ge\frac{2}{3}\left(a+b+c\right)\)
từ (1)→\(2\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\ge\frac{2}{3}\left(a+b+c\right)\)
↔ \(S\ge\frac{1}{3}\left(a+b+c\right)=1\)(đặt S luôn cho tiện)
dấu = xảy ra khi BĐt ở đầu đúng :\(\begin{cases}a=b\\b=c\\c=a\end{cases}\)mà a+b+c=3↔a=b=c=1
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)