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Hình như thiếu điều kiện a;b;c không âm
Áp dụng BĐT Cauchy cho 2 số: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2ab+2bc+2ca\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Tiếp tục dùng Cauchy: \(ab+bc\ge2\sqrt{ab^2c}\)
Tương tự: \(bc+ca\ge2\sqrt{abc^2};ca+ab\ge2\sqrt{a^2bc}\)
\(\Rightarrow2\left(ab+bc+ca\right)\ge2\sqrt{a^2bc}+2\sqrt{ab^2c}+2\sqrt{abc^2}\)
\(\Leftrightarrow ab+bc+ca\ge\sqrt{a^2bc}+\sqrt{ab^2c}+\sqrt{abc^2}=\sqrt{abc}.\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrow a^2+b^2+c^2\ge\sqrt{abc}.\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
Dấu "=" xảy ra <=> a=b=c.
\(\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
thiếu a;b;c k âm nhá
áp dụng bđt cosi ngược ta có:
\(\sqrt{bc}< =\frac{b+c}{2}\Rightarrow a\sqrt{bc}< =\frac{a\left(b+c\right)}{2}=\frac{ab+ac}{2}\)
tương tự \(b\sqrt{ac}< =\frac{ab+bc}{2};c\sqrt{ab}< =\frac{ac+bc}{2}\)
\(\Rightarrow\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}< =\frac{ab+ac}{2}+\frac{ab+bc}{2}+\frac{ac+bc}{2}\)
\(=\frac{ab+ac+ab+bc+ac+bc}{2}=\frac{2ab+2bc+2ac}{2}=ab+ac+bc\left(1\right)\)
\(2ab< =a^2+b^2;2bc< =b^2+c^2;2ac< =a^2+c^2\Rightarrow2ab+2bc+2ac< =\)
\(a^2+b^2+b^2+c^2+a^2+c^2=2a^2+2b^2+2c^2\Rightarrow ab+bc+ac< =a^2+b^2+c^2\left(2\right)\)
từ \(\left(1\right)\left(2\right)\Rightarrow\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)< =a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2>=\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(đpcm\right)\)
dấu = xảy ra khi a=b=c
Câu 3:
bạn cứ áp dụng cái \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Câu 4:
từ giả thiết :\(a+b+c+\sqrt{abc}=4\Leftrightarrow\sqrt{abc}=4-a-b-c\Leftrightarrow abc=\left(4-a-b-c\right)^2\)
ta có: \(a\left(4-b\right)\left(4-c\right)=a\left(16-4c-4b+bc\right)=16a-4ac-4ab+abc\)
\(=16a-4ab-4ac+\left[4-\left(a+b+c\right)\right]^2=16a-4ab-4ac+16-8\left(a+b+c\right)+\left(a+b+c\right)^2\)
\(=a^2+b^2+c^2-2ab-2ac+2bc+8a-8b-8c+16\)
\(=\left(a-b-c\right)^2+8\left(a-b-c\right)+16=\left(a-b-c+4\right)^2\)
\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=a-b-c+4\)(vì \(a-b-c+4=a-b-c+a+b+c+\sqrt{abc}=2a+\sqrt{abc}>0\))
các căn thức còn lại tương tự ...
Ta có :(a+b-c)2 \(\ge\) 0
<=>a2+b2+c2 \(\ge\) 2(bc-ab+ac)
<=>\(\frac{5}{3}\ge\) 2(bc-ab+ac)
<=>bc+ac-ab \(\le\frac{5}{6}< 1\)
<=>\(\frac{bc+ac-ab}{abc}< \frac{1}{abc}\) (vì a,b,c>0 nên chia cả 2 vế cho abc)
<=>\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< 1\) (đpcm)
Giả thiết có: abc+bca+cda+dab = a+b+c+d+\(\sqrt{2012}\)
\(\Leftrightarrow\) (abc+bca+cda+dab-a-b-c-d)2 =2012
\(\Leftrightarrow\) \(\left[\left(abc-c\right)+\left(dab-d\right)+\left(bcd-b\right)+\left(cda-a\right)\right]^2\) = 2012
\(\Leftrightarrow\) \(\left[c\left(ab-1\right)+d\left(ab-1\right)+b\left(cd-1\right)+a\left(cd-1\right)\right]^2\) = 2012
\(\Leftrightarrow\) \(\left[\left(ab-1\right)\left(c+d\right)+\left(ab-1\right)\left(a+b\right)\right]^2\) = 2012
Áp dụng BĐT Bunhia cho 2 cặp số: (ab-1 ; a+b);(cd-1 ; c+d)
Ta có: \(\left[\left(ab-1\right)\left(c+d\right)+\left(ab-1\right)\left(a+b\right)\right]^2\) \(\le\) \(\left[\left(ab-1\right)^2+\left(a+b\right)^2\right]\left[\left(cd-1\right)^2+\left(c+d\right)^2\right]\)
\(\Leftrightarrow\) 2012 \(\le\) ( a2b2-2ab+1+a2+2ab+b2) (c2d2-2cd+1+c2+2cd+d2)
\(\Leftrightarrow\) 2012\(\le\) ( a2b2 +a2+b2+1)(c2d2+c2+d2+1)
\(\Leftrightarrow\) 2012 \(\le\) (a2+1)(b2+1)(c2+1)(d2+1) (đpcm)