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Ta có : \(a^2+b^2\ge2ab\Rightarrow a^2+b^2-ab\ge ab\)
\(\Rightarrow\dfrac{1}{a^2-ab+b^2}\le\dfrac{1}{ab}=\dfrac{abc}{ab}=c\) ( do $abc=1$ )
Tương tự ta có :
\(\dfrac{1}{b^2-bc+c^2}\le a\)
\(\dfrac{1}{c^2-ab+a^2}\le b\)
Cộng vế với vế các BĐT trên có :
\(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
Dấu "=" xảy ra khi $a=b=c$
\(VT=\dfrac{1}{a^2+b^2-ab}+\dfrac{1}{b^2+c^2-bc}+\dfrac{1}{c^2+a^2-ca}\)
\(VT\le\dfrac{1}{2ab-ab}+\dfrac{1}{2bc-bc}+\dfrac{1}{2ca-ca}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Do: \(a^2+b^2+c^2=1\text{ nen }a^2\le1,b^2\le1,c^2\le1\)
\(\Rightarrow a\ge-1;b\ge-1;c\ge-1\)
\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge0\)
\(\Rightarrow1+a+b+c+ab+bc+ca+abc\ge0\)
Cần C/m:
\(1+a+b+c+ab+bc+ca\ge0\)
Ta có:
\(1+a+b+c+ab+bc+ca\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+ab+bc+ca+a+b+c\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2+2\left(a+b+c\right)+2ab+2bc+2ca+abc\ge0\)
\(\Leftrightarrow\left(a+b+c\right)^2+2\left(a+b+c\right)+1\ge0\)
\(\Leftrightarrow\left(a+b+c+1\right)^2\ge0\left(\text{luon dung}\right)\)
=> ĐPCM
\(a^2+b^2+c^2=1\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=1+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2=1+2\left(ab+bc+ca\right)\)
\(\Rightarrow1+2\left(ab+bc+ca\right)\ge0\)
\(\Rightarrow ab+bc+ca\ge-\dfrac{1}{2}\)
Lại có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow ab+bc+ca\le a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca\le1\)
Ta có : \(a^2+ab=c^2+bc\Leftrightarrow a^2-c^2+b\left(a-c\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left(a+b+c\right)=0\Leftrightarrow a-c=0\) ( do a;b;c \(\ne0\Rightarrow a+b+c\ne0\) )
\(\Leftrightarrow a=c\)
Làm tương tự ; ta có : a = b . Suy ra : a = b = c
\(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=6\)
Vậy ...
Ta có : a2+ab=c2+bc⇔a2−c2+b(a−c)=0a2+ab=c2+bc⇔a2−c2+b(a−c)=0
⇔(a−c)(a+b+c)=0⇔a−c=0⇔(a−c)(a+b+c)=0⇔a−c=0 ( do a;b;c ≠0⇒a+b+c≠0≠0⇒a+b+c≠0 )
⇔a=c⇔a=c
Làm tương tự ; ta có : a = b . Suy ra : a = b = c
A=(1+ab)(1+bc)(1+ca)=(1+1)(1+1)(1+1)=6A=(1+ab)(1+bc)(1+ca)=(1+1)(1+1)(1+1)=6
Vậy ...
Ta có: \(0\le a\le b\le1.\)
\(\Rightarrow\left\{{}\begin{matrix}a-1\le0\\b-1\le0\end{matrix}\right.\)
\(\Rightarrow\left(a-1\right).\left(b-1\right)\ge0\)
\(\Rightarrow ab-a-b+1\ge0.\)
\(\Rightarrow ab+1\ge0+a+b\)
\(\Rightarrow ab+1\ge a+b\)
\(\Rightarrow\frac{1}{ab+1}\le\frac{1}{a+b}.\)
\(\Rightarrow\frac{c}{ab+1}\le\frac{c}{a+b}\left(c\ge0\right).\)
Mà \(\frac{c}{a+b}\le\frac{2c}{a+b+c}\left(c\ge0\right)\)
\(\Rightarrow\frac{c}{ab+1}\le\frac{2c}{a+b+c}\left(1\right).\)
Chứng minh tương tự ta cũng có:
\(\frac{b}{ac+1}\le\frac{2b}{a+b+c}\left(2\right);\frac{a}{bc+1}\le\frac{2a}{a+b+c}\left(3\right).\)
Cộng theo vế \(\left(1\right);\left(2\right)và\left(3\right)\) ta được:
\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}\)
\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a+2b+2c}{a+b+c}\)
\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2.\left(a+b+c\right)}{a+b+c}\)
\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\left(đpcm\right).\)
Chúc bạn học tốt!
Cảm ơn bạn nha!