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xét hiệu
\(\dfrac{a^2+b^2+c^2}{3}-\dfrac{\left(a+b+c\right)^2}{9}\ge0\)
<=> \(\dfrac{3\left(a^2+b^2+c^2\right)}{9}-\dfrac{\left(a+b+c\right)^2}{9}\ge0\)
<=>\(3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc\ge0\)
<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)
<=> (a-b)2 +(b-c)2 +(c-a)2 ≥ 0 (luôn đúng)
=> đpcm)
b. \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c\ge0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
-Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Xét hiệu VT - VP
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ab+b^2}+\dfrac{c+a}{ab+c^2}-\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}=\dfrac{a^2+ab-bc-a^2}{a\left(bc+a^2\right)}+\dfrac{b^2+bc-ac-b^2}{b\left(ac+b^2\right)}+\dfrac{c^2+ac-ab-c^2}{c\left(ab+c^2\right)}=\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}\)
Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0
\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)
=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)
mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm
\(a^2+b^2+c^2+1>a+b+c\)
\(\Leftrightarrow a^2+b^2+c^2+1-a-b-c>0\)
\(\Leftrightarrow\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)+\frac{1}{4}>0\)
\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2+\frac{1}{4}>0\)( luôn đúng )
Vậy ...
Ta có: \(a^2+b^2+c^2+1>a+b+c\)
\(\Leftrightarrow a^2+b^2+c^2+1-a-b-c>0\)
\(\Leftrightarrow\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)+\frac{1}{4}>0\)
\(\Leftrightarrow\left(a^2-2.a.\frac{1}{2}+\frac{1}{4}\right)+\left(b^2-2.b.\frac{1}{2}+\frac{1}{4}\right)+\left(c^2-2.c.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}>0\)
\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2+\frac{1}{4}>0\)
Ta thấy: (a-1/2)2 lớn hơn hoặc bằng 0 (với mọi a)
(b-1/2)2 lớn hơn hoặc bằng 0 (với mọi b)
(c-1/2)2 lớn hơn hoặc bằng 0 (với mọi c)
1/4 > 0
Nên BĐT luôn đúng
=> ĐPCM
\(a^2+b^2+c^2+1>a+b+c\)
\(\Leftrightarrow\left(a^2-a+\dfrac{1}{4}\right)+\left(b^2-b+\dfrac{1}{4}\right)+\left(c^2-c+\dfrac{1}{4}\right)+\dfrac{1}{4}>0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}\right)^2+\left(b-\dfrac{1}{2}\right)^2+\left(c-\dfrac{1}{2}\right)^2+\dfrac{1}{4}>0\) (đúng)
\(\Rightarrow\)ĐPCM
Ta chứng minh:
a2 + b2 + c2 \(\ge\) ab + bc + ac
Nhân cả 2 vế với 2 ta được :
= 2a2 + 2b2 + 2c2 \(\ge\) 2ab + 2bc + 2ac
= 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac \(\ge0\)
= ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( a2 - 2ac + c2 ) \(\ge0\)
= ( a - b )2 + ( b - c)2 + ( a - c )2 \(\ge0\) ( luôn đúng )
\(\Rightarrow\)a2 + b2 + c2 \(\ge\)ab + bc + ac
Ta có : a2 + b2 + c2 \(\ge\)ab + bc + ac
Nhân cả 2 vế với 2 ta được :
2 ( a2 + b2 + c2 ) \(\ge\) 2 ( ab + bc + ac )
Cộng cả 2 vế với : a2 + b2 + c2 ta được :
3 ( a2 + b2 + c2 ) \(\ge\) a2 + b2 + c2 + 2ab + 2bc + 2ac
3 ( a2 + b2 + c2 ) \(\ge\) ( a + b + c )2
3 ( a2 + b2 + c2 ) \(\ge\)1
a2 + b2 + c2 \(\ge\)\(\frac{1}{3}\) ( đpcm)
\(a^2+b^2+c^2\ge\frac{1}{3}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đung)
Xét hiệu : \(\left(a^2+b^2+c^2+1\right)-\left(a+b+c\right)\)
\(=\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)+\frac{1}{4}\)
\(=\left(c-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2+\frac{1}{4}>0\forall a,b,c\)
\(\Rightarrow a^2+b^2+c^2+1>a+b+c\left(đpcm\right)\)
cách khác
a2 + b2 + c2 + 1 > a + b + c
<=> 4( a2 + b2 + c2 + 1 ) > 4( a + b + c )
<=> 4a2 + 4b2 + 4c2 + 4 > 4a + 4b + 4c
<=> 4a2 + 4b2 + 4c2 + 4 - 4a - 4b - 4c > 0
<=> ( 4a2 - 4a + 1 ) + ( 4b2 - 4b + 1 ) + ( 4c2 - 4c + 1 ) + 1 > 0
<=> ( 2a - 1 )2 + ( 2b - 1 )2 + ( 2c - 1 )2 + 1 > 0 ( luôn đúng )
Vậy bđt ban đầu được chứng minh