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1a)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+b+a\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi x=y=1
b)\(a^2+b^2+c^2\ge a\left(b+c\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+b^2+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi a=b=c=0
Giả sử \(a\ge b\ge c\)
Ta có:\(\frac{a+b}{ab+c^2}+\frac{b+c}{bc+a^2}+\frac{c+a}{ca+b^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{ac+bc-ab-c^2}{c\left(ab+c^2\right)}+\frac{ab+ac-bc-a^2}{\left(bc+a^2\right)a}+\frac{cb+ab-ca-b^2}{b\left(ca+b^2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}+\frac{\left(c-b\right)\left(b-a\right)}{b\left(ca+b^2\right)}\le0\)
Ta có:\(\left(c-b\right)\left(b-a\right)\ge0;\left(b-a\right)\left(a-c\right)\le0;\left(a-c\right)\left(c-b\right)\le0\)
\(\Rightarrow\frac{\left(c-b\right)\left(c-a\right)}{b\left(ca+b^2\right)}\le\frac{\left(c-b\right)\left(c-a\right)}{c\left(ab+c^2\right)}\)
\(\Rightarrow LHS\le\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(c-b\right)\left(b-a\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\)
\(=\frac{-\left(c-b\right)^2}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)c}\le0\)
\(\Rightarrowđpcm\)
\(0\le a,b,c\le1\Rightarrow b\ge b^2;c\ge c^3\)
\(\Rightarrow a+b^2+c^3\le a+b+c\)
\(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0\)
\(\Leftrightarrow\left(1-b-a+ab\right)\left(1-c\right)\ge0\)
\(\Leftrightarrow1-\left(a+b+c\right)+ab+bc+ca-abc\ge0\)
\(\Leftrightarrow a+b+c-ab-bc-ca\le1-abc\le1\)
=> đpcm
a^2 + b^2 + c^2= ab + bc + ca
2 ( a^2 + b^2 + c^2 ) = 2 ( ab + bc + ca)
2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca
a^2 + a^2 + b^2 + b^2 + c^2+ c^2 – 2ab – 2bc – 2ca = 0
a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c² + a² – 2ca = 0
(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca) = 0
(a – b)^2 + (b – c)^2 + (c – a)^2 = 0
Vì (a-b)^2 lớn hơn hoặc bằng 0 với mọi a và b
(b-c)^2 lớn hơn hoặc bằng 0 với mọi c và b
(c-a)^2 lớn hơn hoặc bằng 0 với mọi a và c
=> (a-b)^2 =0 ; (b-c)^2=0 ; (c-a)^2=0
=> a=b ; b=c ; c=a
=>a=b=c
Bài làm
a) Ta có: ( a - b + c )2 = [ a - ( b - c ) ]2
= a2 - 2a( b - c ) + ( b - c )2
= a2 - 2ab + 2ac + b2 - 2bc + c2
= a2 + b2 + c2 + 2ac - 2ab - 2bc
Mik làm mấy lần rồi nhưng vẫn ra kết quả như vậy, bạn xem lại đề nhé.
b) Ta có: a2 + b2 + c2 > ab + bc + ca
=> 2( a2 + b2 + c2 ) > 2( ab + bc + ca )
=> 2a2 + 2b2 + 2c2 > 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca > 0
=> ( a2 + b2 + c2 ) + ( a2 + b2 + c2 - 2ab - 2bc - 2ca ) > 0
=> ( a2 + b2 + c2 ) + ( a - b - c )2 > 0 ( Luôn đúng )
Vậy a2 + b2 + c2 > ab + bc + ca ( đpcm ).
c) a2 + b2 + 1 > a + b + ab ( mik nghĩ cái a ở vế phải phải là a thôi chứ không phỉa a^2. bạn kiểm tra đề nha )
=> 2a2 + 2b2 + 2 > 2a + 2b + 2ab
=> 2a2 + 2b2 + 2 - 2a - 2b - 2ab > 0
=> ( a2 - 2ab + b2 ) + ( a2 - 2a + 1 ) + ( b2 - 2b + 1 ) > 0
=> ( a - b )2 + ( a - 1 )2 + ( b - 1 )2 > 0 ( luôn đúng )
Vậy a2 + b2 + 1 > a + b + ab ( đpcm )
\(1,\left(a-b+c\right)^2=\left[\left(a-b\right)+c\right]^2\)
\(=\left(a-b\right)^2+2\left(a-b\right)c+c^2\)
\(=a^2+b^2+c^2-2ab-2bc-2ca\)
\(2,..2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Dấu "=" xảy ra khi a = b = c
3, Sửa đề : \(a^2+b^2+1\ge a+b+ab\)
Ta có : \(2a^2+2b^2+2-2a-2b-2ab\)
\(=\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\)
\(=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
\(\Rightarrow2a^2+2b^2+2\ge2a+2b+2ab\)
\(\Leftrightarrow a^2+b^2+1\ge a+b+ab\)
Dấu "=" xảy ra khi a = b = 1
Other way:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)
Dấu "=" xảy ra khi a=b=c
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)
\(\ge\left(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\) \(\left(1\right)\)
Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\) ( Do abc=1 )
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)
Mà \(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\) (đpcm)
Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:
\(\left\{{}\begin{matrix}\dfrac{a+b}{2}\ge\sqrt{ab}\\\dfrac{b+c}{2}\ge\sqrt{bc}\\\dfrac{a+c}{2}\ge\sqrt{ac}\end{matrix}\right.\)
Cộng theo 3 vế ta có:
\(\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{a+c}{2}\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}b+\dfrac{1}{2}c+\dfrac{1}{2}a+\dfrac{1}{2}c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\left(đpcm\right)\)
\(a=b=c\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+c^2=2bc\\a^2+c^2=2ac\end{matrix}\right.\)
Cộng theo 3 vế ta có:
\(a^2+b^2+b^2+c^2+a^2+c^2=2ab+2bc+2ac\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
Ngược lại,khi \(a\ne b\ne c\) thì \(\left\{{}\begin{matrix}a^2+b^2>2ab\\b^2+c^2>2bc\\a^2+c^2>2ac\end{matrix}\right.\) ta có thể dễ dàng cm được \(a^2+b^2+c^2>ab+bc+ac\)
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( Bất đẳng thức luôn đúng )
Vậy \(a^2+b^2+c^2\ge ab+bc+ca\forall a;b;c\)
Tham khảo nhé~
cách khác:
\(\left(a-b\right)^2\ge0\)
<=> \(a^2-2ab+b^2\ge0\)
<=> \(a^2+b^2\ge2ab\)
Tương tự: \(b^2+c^2\ge2bc;\)\(c^2+a^2\ge2ca\)
Cộng theo vế ta được:
\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
<=> \(a^2+b^2+c^2\ge ab+bc+ca\) \(\forall a,b,c\)