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Áp dụng BĐT Cauchy dạng phân thức :
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\ge\frac{9}{ab+bc+ac}\)
\(\Rightarrow VT\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ac}\)
\(\Leftrightarrow VT\ge\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{7}{ab+ac+bc}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
\(\Rightarrow\frac{7}{ab+bc+ac}\ge21\left(1\right)\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}\)
\(\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=9\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge21+9=30\left(đpcm\right)\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}\)
\(\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=9+\frac{7}{\frac{1}{3}}=30\)
Theo bất đẳng thức Cauchy dạng phân thức
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}>\frac{9}{ab+bc+ac}.\)
\(VT>\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ac}\)
\(VT>\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{7}{ab+bc+ac}\)
Theo hệ quả của bất đẳng thức Cauchy
\(ab+bc+ac< \frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
\(\frac{7}{ab+bc+ac>21}\left(1\right)\)
Theo bất đẳng thức Cauchy dạng phân thức
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}>\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)
Từ (1) và (2)
\(VT>21+9=30\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=\frac{1}{3}\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Trước khi đọc lời giải hãy thăm nhà em trước nhé ! See method from solution! Cảm ơn mn!
Ok, giờ chú ý:
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{ab.ca+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=1\) với abc = 1.
Như vậy: \(VT=\sqrt{\left(\Sigma\frac{1}{\sqrt{ab+a+2}}\right)^2}\le\sqrt{3\left(\Sigma\frac{1}{\frac{\left(ab+a+1\right)}{3}+\frac{\left(ab+a+1\right)}{3}+\frac{\left(ab+a+1\right)}{3}+1}\right)}\)
\(\le\sqrt{\frac{3}{16}\left[\Sigma\left(\frac{9}{ab+a+1}+1\right)\right]}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
\(ab+bc+ca=1\Leftrightarrow ab+bc+ca+a^2=1+a^2\)
\(\Leftrightarrow1+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự ta có: \(1+b^2=\left(b+a\right)\left(b+c\right),1+c^2=\left(c+a\right)\left(c+b\right)\)
Suy ra:
\(\frac{a-b}{1+c^2}+\frac{b-c}{1+a^2}+\frac{c-a}{1+b^2}=\frac{a-b}{\left(c+a\right)\left(c+b\right)}+\frac{b-c}{\left(a+b\right)\left(a+c\right)}+\frac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\frac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Không mất tính tổng quát giả sử \(c=max\left\{a,b,c\right\}\)
\(\Rightarrow2c\ge a+b\)
\(\Rightarrow c\ge\frac{a+b}{2}\)
Từ giả thiết \(\Rightarrow a,b\le1\)
\(\Rightarrow ab\le1\)( *)
Đặt \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{5}{2}\)
\(=\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}\)
Đặt \(S=\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}-\frac{5}{2}\)
Xét hiệu \(P-S=\)\(\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}-\)\(-\frac{1}{a+b+\frac{1}{a+b}}-a-b-\frac{1}{a+b}+\frac{5}{2}\)
\(=\frac{1}{\frac{ab+b^2+1-ab}{a+b}}+\frac{1}{\frac{a^2+ab+1-ab}{a+b}}-\frac{1}{\frac{\left(a+\right)^2+1}{a+b}}-\left(a+b\right)\)
\(=\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)
Ta sẽ chứng minh \(\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\ge0\)
\(\Leftrightarrow\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}\ge\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)
\(\Leftrightarrow\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge1+\frac{1}{1+\left(a+b\right)^2}\)
\(\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2+\left(a+b\right)^2}{1+\left(a+b\right)^2}\)
\(\Rightarrow\left(2+b^2+a^2\right)\left[1+\left(a+b\right)^2\right]\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2\right)\left(1+b^2\right)\)
\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2+b^2+a^2b^2\right)\)
\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2-2a^2b^2-\left(a+b\right)^2\left(a^2+b^2\right)-\left(a+b\right)^2a^2b^2\)\(-2-2\left(a^2+b^2\right)-\left(a+b^2\right)\ge0\)
\(\Leftrightarrow-2a^2b^2-\left(a+b\right)^2a^2b^2+a^2+b^2-\left(a+b\right)^2\ge0\)
\(\Leftrightarrow ab\left[ab\left(a+b\right)^2+2ab-2\right]\le0\)
\(\Leftrightarrow ab\left(a+b\right)^2+2ab-2\le0\)( do a,b \(\ge0\))
\(\Leftrightarrow ab\left(a+b\right)^2\le2\left(1-ab\right)\)
\(\Leftrightarrow ab\left(a+b\right)^2\le2c\left(a+b\right)\) (1)
Mà \(c\ge\frac{a+b}{2}\)
\(\Rightarrow2c\left(a+b\right)\ge\left(a+b\right)^2\)
Ta có: \(\left(a+b\right)^2\ge ab\left(a+b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2\left(1-ab\right)\ge0\)( đúng do (*) )
\(\Rightarrow\left(1\right)\)đúng
\(\Rightarrow P-S\ge0\)
\(\Rightarrow P\ge S\)
Ta phải chứng minh \(S\ge0\)
\(\Leftrightarrow\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}\ge\frac{5}{2}\)
\(\Leftrightarrow\frac{a+b}{1+\left(a+b\right)^2}+\frac{1+\left(a+b\right)^2}{a+b}\ge\frac{5}{2}\) (2)
Đặt \(x=\frac{1+\left(a+b\right)^2}{a+b}\)
Ta có: \(1+\left(a+b\right)^2\ge2\left(a+b\right)\)
\(\Leftrightarrow\left(a+b-1\right)^2\ge0\)( đúng )
\(\Rightarrow x=\frac{1+\left(a+b\right)^2}{a+b}\ge2\)
=> (2) có dạng \(x+\frac{1}{x}\ge\frac{5}{2}\)
\(\Leftrightarrow2x^2-5x+2\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\ge0\)( đúng )
\(\Rightarrow S\ge0\)mà \(P\ge S\)
\(\Rightarrow P\ge0\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{5}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab+bc+ca=1\\ab\left[ab\left(a+b\right)^2+2ab-2\right]=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=c=1;b=0\\b=c=1;a=0\end{cases}}\)
Ta có \(\frac{a.1-bc}{a.1+bc}==\frac{a^2+ac}{a^2+ab+bc+ca}=\frac{a}{a+b}\)
Từ đó \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)
\(=-\left(\frac{a}{c-1}+\frac{b}{a-1}+\frac{c}{b-1}\right)=-\left(\frac{a^2}{ca-a}+\frac{b^2}{ab-b}+\frac{c^2}{bc-c}\right)\)
\(\le-\frac{\left(a+b+c\right)^2}{ab+bc+ca-\left(a+b+c\right)}=-\frac{1}{ab+bc+ca-1}\le-\frac{1}{\frac{\left(a+b+c\right)^2}{3}-1}=\frac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}.\)
Lời giải:
Áp dụng hệ quả quen thuộc của BĐT AM-GM:
$3(ab+bc+ac)\leq (a+b+c)^2$
$1=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{27}$
Do đó:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ac}\geq \frac{(1+1+1+1)^2}{a^2+b^2+c^2+3(ab+bc+ac)}=\frac{16}{(a+b+c)^2+ab+bc+ac}\)
\(\geq \frac{16}{(a+b+c)^2+\frac{(a+b+c)^2}{3}}=\frac{12}{(a+b+c)^2}=12\)
\(\frac{2}{3ab}+\frac{2}{3bc}+\frac{2}{3ac}=\frac{2}{3}.\frac{a+b+c}{abc}=\frac{2}{3abc}\geq \frac{2}{3.\frac{1}{27}}=18\)
Cộng 2 BĐT trên lại:
\(\Rightarrow \frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\geq 12+18=30\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Lời giải:
Áp dụng hệ quả quen thuộc của BĐT AM-GM:
$3(ab+bc+ac)\leq (a+b+c)^2$
$1=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{27}$
Do đó:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ac}\geq \frac{(1+1+1+1)^2}{a^2+b^2+c^2+3(ab+bc+ac)}=\frac{16}{(a+b+c)^2+ab+bc+ac}\)
\(\geq \frac{16}{(a+b+c)^2+\frac{(a+b+c)^2}{3}}=\frac{12}{(a+b+c)^2}=12\)
\(\frac{2}{3ab}+\frac{2}{3bc}+\frac{2}{3ac}=\frac{2}{3}.\frac{a+b+c}{abc}=\frac{2}{3abc}\geq \frac{2}{3.\frac{1}{27}}=18\)
Cộng 2 BĐT trên lại:
\(\Rightarrow \frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\geq 12+18=30\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Áp dụng Cauchy Schwarz ta dễ có:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(=\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\right)+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=30\)
Đẳng thức xảy ra tại a=b=c=1/3
giúp em hiểu chỗ \(\frac{7}{ab+bc+ca}\Rightarrow\frac{7}{\frac{\left(a+b+c\right)^2}{3}}\)