Chứng minh đẳng thức:
\(\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Chứng minh
\(\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(A=\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}\)
\(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
..................................
\(A=\dfrac{a^2+b^2-2ab}{ab}+\dfrac{b^2-2ab+c^2}{bc}+c^2+a^2-\dfrac{2ca}{ca}\)
\(A=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}-2\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}-2\right)=\dfrac{\left(b+c\right)}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}-6\)
\(A=\left[\dfrac{\left(b+c\right)}{a}+1\right]+\left[\dfrac{\left(a+c\right)}{b}+1\right]+\left[\dfrac{\left(a+b\right)}{c}+1\right]-9\)
\(A=\dfrac{\left(a+b+c\right)}{a}+\dfrac{\left(a+b+c\right)}{b}+\left[\dfrac{\left(a+b+c\right)}{c}\right]-9\)
\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-9\)
Ket luan
\(A\ne B\) => đề sai--> hoặc mình công trừ sai
Cho a,b,c>0 và a+b+c=3Chứng minh \(\dfrac{a\left(a+bc\right)^2}{b\left(ab+2c^2\right)}+\dfrac{b\left(b+ca\right)^2}{c\left(bc+2a^2\right)}+\dfrac{c\left(c+ab\right)^2}{a\left(ca+2b^2\right)}\ge4\)
\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)}{a^2c^2+2ab^2c}\)
\(P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)
\(P\ge\dfrac{\left[a^2+b^2+c^2+3abc\right]^2}{\left(ab+bc+ca\right)^2}\)
Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)
Ta có: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)
\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)
\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)
sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Cho ba số a, b, c thỏa mãn điều kiện: \(\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0\)
Chứng minh rằng: \(\dfrac{a}{\left(bc-a^2\right)^2}+\dfrac{b}{\left(ca-b^2\right)^2}+\dfrac{c}{\left(ab-c^2\right)^2}=0\)
Chứng minh bất đẳng thức sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(a,b,c>0\right)\)
Áp dụng BĐT cosi:
\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)
Dấu \("="\Leftrightarrow a=b=c\)
Cho a,b,c>0 thỏa mãn : \(ab+bc+ca=0\)
C/m: \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3+\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\dfrac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\dfrac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
may cai nay tuong hoi truoc co nguoi dang roi ma
ta có:
\(\sqrt{\dfrac{\left(a+b\right).\left(a+c\right)}{a^2}}\le\dfrac{1}{2}.\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)=a+\dfrac{b}{2}+\dfrac{c}{2}\)
tương tự thì ta có:
\(VP\le3+2\left(a+b+c\right)\)
\(VP=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}\)
từ các điều trên ta thấy cần CM:
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge a+b+c\)
bạn tự CM nốt ạ
cho a,b,c>0 chứng minh rằng
\(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\ge\dfrac{4}{ab+bc+ca}\)
Fix đề: Cho a,b,c không âm. Chứng minh \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\ge\dfrac{4}{ab+bc+ca}\)
Dự đoán điểm rơi sẽ có 1 số bằng 0.
Giả sử \(c=min\left\{a,b,c\right\}\) ( c là số nhỏ nhất trong 3 số) thì \(c\ge0\)
do đó \(ab+bc+ca\ge ab\) và \(\dfrac{1}{\left(b-c\right)^2}\ge\dfrac{1}{b^2};\dfrac{1}{\left(c-a\right)^2}=\dfrac{1}{\left(a-c\right)^2}\ge\dfrac{1}{a^2}\)
BDT cần chứng minh tương đương
\(ab\left[\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\ge4\)
\(\Leftrightarrow\dfrac{ab}{\left(a-b\right)^2}+\dfrac{a^2+b^2}{ab}\ge4\)
\(\Leftrightarrow\dfrac{ab}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{ab}+2\ge4\)
BĐT trên hiển nhiên đúng theo AM-GM.
Do đó ta có đpcm. Dấu = xảy ra khi c=0 , \(\left(a-b\right)^2=a^2b^2\) ( và các hoán vị )
Cho \(a,b,c\) thỏa mãn \(\left|a\right|,\left|b\right|,\left|c\right|< 1\) và \(ab+bc+ca=2\). Chứng minh :
\(P=\dfrac{a^2}{1-b^2}+\dfrac{b^2}{1-c^2}+\dfrac{c^2}{1-a^2}\ge6\).
\(a^2+b^2+c^2\ge ab+bc+ca=2\)
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 \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)