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\(a^2+b^2+c^2\ge ab+bc+ca=2\)
Áp dụng BĐT C-S:
\(P\ge\dfrac{\left(a+b+c\right)^2}{3-\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2+4}{3-\left(a^2+b^2+c^2\right)}\)
Đặt \(a^2+b^2+c^2=x\)
Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)
\(\Leftrightarrow x\ge2\) (đúng)
Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)
\((\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0 \)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]}+\dfrac{1}{\left(c-a\right)\left[\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\right]}+\dfrac{1}{\left(a-b\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]}=0\)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\dfrac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\dfrac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)(t/m)
Suy ra ta được Đt cần chứng minh.
Chúc bạn học tốt với hoc24 nha
Lời giải:
Ta có:
\(\frac{1}{(b-c)(a^2+ac-b^2-bc)}+\frac{1}{(c-a)(b^2+bc-c^2-ca)}+\frac{1}{(a-b)(c^2+cb-a^2-ab)}\)
\(=\frac{1}{(b-c)[(a^2-b^2)+(ac-bc)]}+\frac{1}{(c-a)[(b^2-c^2)+(ba-ca)]}+\frac{1}{(a-b)[(c^2-a^2)+(cb-ab)]}\)
\(=\frac{1}{(b-c)[(a-b)(a+b)+c(a-b)]}+\frac{1}{(c-a)[(b-c)(b+c)+a(b-c)]}+\frac{1}{(a-b)[(c-a)(c+a)+b(c-a)]}\)
\(=\frac{1}{(b-c)(a-b)(a+b+c)}+\frac{1}{(c-a)(b-c)(b+c+a)}+\frac{1}{(a-b)(c-a)(c+a+b)}\)
\(=\frac{(c-a)+(a-b)+(b-c)}{(a-b)(b-c)(c-a)(a+b+c)}=\frac{0}{(a-b)(b-c)(c-a)(a+b+c)}=0\)
Ta có đpcm.
\(A=\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}\)
\(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
..................................
\(A=\dfrac{a^2+b^2-2ab}{ab}+\dfrac{b^2-2ab+c^2}{bc}+c^2+a^2-\dfrac{2ca}{ca}\)
\(A=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}-2\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}-2\right)=\dfrac{\left(b+c\right)}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}-6\)
\(A=\left[\dfrac{\left(b+c\right)}{a}+1\right]+\left[\dfrac{\left(a+c\right)}{b}+1\right]+\left[\dfrac{\left(a+b\right)}{c}+1\right]-9\)
\(A=\dfrac{\left(a+b+c\right)}{a}+\dfrac{\left(a+b+c\right)}{b}+\left[\dfrac{\left(a+b+c\right)}{c}\right]-9\)
\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-9\)
Ket luan
\(A\ne B\) => đề sai--> hoặc mình công trừ sai
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)
Lại có:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)
\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)
\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)
=> \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)
\(=a\left(ab+ca\right)+b+c\) (Vì ab+bc+ca=1)
\(=\left(a^2+1\right)\left(b+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (Vì \(a^2+1=\left(a+b\right)\left(c+a\right)\))
\(T=1\)
Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
nhân cả vế với abc ta có điều cần chứng minh
\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)
VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)
=>(đpcm)
mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
chỉ cần chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé
