\(\frac{a^2}{a^2+ab+b^2}\)+\(\frac{b^2}{b^2+bc+c^2}\)+\(\frac{c^2}{c^2+ac+a^2}\)≥1 với mọi a;b;c là các số thực dương
Những câu hỏi liên quan
Mọi người giúp em bài này với ạ
Em cảm ơn ạ
Cho a+b+c=1
Chứng minh: \(\frac{c+ab}{a^2+b^2+abc-1}+\frac{a+bc}{b^2+c^2+abc-1}+\frac{b+ac}{a^2+c^2+abc-1}=\frac{bc+ab+ac+8}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)
frac{a^2b+bc^2-1}{acleft(a+cright)}+frac{b^2c+ca^2-1}{ableft(a+bright)}+frac{c^2a+ab^2-1}{bcleft(b+cright)}frac{a^2b^2+b^2c^2-b}{a+c}+frac{b^2c^2+c^2a^2-c}{a+b}+frac{c^2a^2+a^2b^2-a}{b+c}frac{frac{1}{a^2}-frac{1}{ac}+frac{1}{c^2}}{a+c}+frac{frac{1}{b^2}-frac{1}{ab}+frac{1}{a^2}}{a+b}+frac{frac{1}{c^2}-frac{1}{bc}+frac{1}{b^2}}{b+c}gefrac{1}{acleft(a+cright)}+frac{1}{bcleft(b+cright)}+frac{1}{ableft(b+aright)}frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}gefrac{3}{2}
Đọc tiếp
\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
CMR: \(\frac{a^2+b^2+c^2}{ab+bc+ac} + \frac{1}{3} \geq \frac{8}{9}(\frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b})\)
CMR:\((1+a+b+c)(1+ab+bc+ac) \geq 4\sqrt{2(a+bc)(b+ac)(c+ab)}\)
Chứng minh rằng: \(\frac{1}{2}+\frac{a^2+b^2+c^2}{ab+bc+ca}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)với mọi số thực a,b,c
Cho a,b,c là các số dương. CMR \(\frac{ab}{a^2+bc+ca}+\frac{bc}{b^2+ca+ab}+\frac{ca}{c^2+ab+bc}\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)Mọi người giúp em với ạ!
Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)
\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)
\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)
Nhân phá và rút gọn 2 vế:
\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)
Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
Đúng 1
Bình luận (0)
a)chứng minh rằng: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\) với mọi giá trị của a,b
b) cho các số dương a,b,c >0 cmr \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{a+b+c}{3}\)
a/ Biến đổi tương đương:
\(\Leftrightarrow3a^2-3ab+3b^2\ge a^2+ab+b^2\)
\(\Leftrightarrow2\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow2\left(a-b\right)^2\ge0\) (luôn đúng)
b/ \(\frac{a^3}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\frac{a+b}{3}=\frac{2a}{3}-\frac{b}{3}\)
Tương tự: \(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b}{3}-\frac{c}{3}\) ; \(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c}{3}-\frac{a}{3}\)
Cộng vế với vế ta có đpcm
CMR với mọi a, b, c > 0 thì:
\(a^2+b^2+c^2+2abc=1\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)
\(\Leftrightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2=\frac{1}{abc}\)
Đặt \(\left(\frac{a}{bc};\frac{b}{ac};\frac{c}{ab}\right)=\left(x;y;z\right)\)
\(\Rightarrow x+y+z+2=xyz\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\)
\(\Leftrightarrow\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2\)
\(\Leftrightarrow\frac{a}{a+bc}+\frac{b}{c+ca}+\frac{c}{c+ab}=2\)
CMR với mọi a, b, c > 0 thì:
\(a^2+b^2+c^2+2abc=1\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)
\(\Leftrightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2=\frac{1}{abc}\)
Đặt : \(\left(\frac{a}{bc};\frac{b}{ac};\frac{c}{ab}\right)=\left(x,y,z\right)\)
\(x+y+z+2=xyz\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+1=1\)
\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2\)
\(\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)
CMR với mọi a, b, c > 0 thì:
\(a^2+b^2+c^2+2abc=1\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)