Cho 3 số thực dương a, b, c. CMR: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) thì \(\frac{\left(a+b\right)^3}{c^3}+\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}=24\)
Cho 3 số thực dương a,b,c. CMR: \(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(c+a\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
Với x là số dương, áp dụng bđt cauchy ta có:
\(\sqrt{x^3+1}=\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\le\frac{x+1+x^2-x+1}{2}=\frac{x^2+2}{2}\)
=> \(\sqrt{\frac{1}{x^3+1}}\ge\frac{2}{x^2+2}\left(1\right)\)
Áp dụng bđt (1) ta được:
\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}=\sqrt{\frac{1}{1+\left(\frac{b+c}{a}\right)^3}}\ge\frac{2}{\left(\frac{b+c}{a}\right)^2+2}=\frac{2a^2}{\left(b+c\right)^2+2a^2}\)
Suy ra \(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{2a^2}{2\left(b^2+c^2\right)+2a^2}=\frac{a^2}{a^2+b^2+c^2}\left(2\right)\)
Tương tự ta có: \(\sqrt{\frac{b^3}{b^3+\left(c+a\right)^3}}\ge\frac{b^3}{a^3+b^3+c^3}\left(3\right);\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge\frac{c^3}{a^3+b^3+c^3}\left(4\right)\)
Cộng (2),(3),(4) vế theo vế:
\(VT\ge\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Dấu "=" xảy ra khi a=b=c
cho a;b;c là 3 số dương. CMR:
\(\left(a^3+b^3+c^3\right)\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\ge\frac{3}{2}\left(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\right)\)
Cho 3 số thực dương \(a,b,c\) thỏa mãn \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
Chứng minh rằng: \(\left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3=24\)
Lời giải:
Vi $a,b,c>0\Rightarrow a+b+c\neq 0$
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\)
\(\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)
\(\Rightarrow \left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3=2^3+2^3+2^3=24\) (đpcm)
Cho ba số thực a,b,c dương. CMR
\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(c+a\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
Cứu mk !!!!!!!!!!!
Bài này có trong đề thi HSG 9 của huyện hay tỉnh nào đấy :)) được cái thầy t bắt cày đi cày lại cả chục cái đề thi nên bài này t nhớ lắm :))
Với x là số dương, áp dụng bđt Cô-si
\(\sqrt{x^3+1}=\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\le\frac{x+1+x^2-x+1}{2}=\frac{x^2+2}{2}\)
\(\Rightarrow\sqrt{\frac{1}{x^3}}\ge\frac{2}{x^2+2}\) (*)
Dấu (=) xảy ra khi x = 2
Áp dụng bđt (*)
\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}=\sqrt{\frac{1}{1+\left(\frac{b+c}{a}\right)^3}}\ge\frac{2}{\left(\frac{b+c}{a}\right)^2+2}=\frac{2a^2}{\left(b+c\right)^2+2a^2}\)
\(\Rightarrow\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{2a^2}{2\left(b^2+c^2\right)+2a^2}=\frac{a^2}{a^2+b^2+c^2}\left(1\right)\)
CMTT :
\(\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}\ge\frac{b^2}{a^2+b^2+c^2}\) (2)
\(\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge\frac{c^2}{a^2+b^2+c^2}\) (3)
Cộng vế với vế của (1) ; (2) ; (3) ; ta được ĐPCM
Cho a,b,c là 3 số thực dương thỏa mãn abc = 1. CMR:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)
\(VT=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{z^3xy}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3}{2}\sqrt[3]{xyz}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Cho a,b,c là các số thực dương. Chứng minh rằng
\(\left|\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\right|\le\frac{1}{4}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
giả sử \(a\ge b\ge c\ge0\)
Ta có: \(a+\frac{b}{2}-\frac{a^2+ab+b^2}{a+b}=\frac{1}{2}\left(ab-b^2\right)\ge0\Rightarrow a+\frac{b}{2}\ge\frac{a^2+ab+b^2}{a+b}\)
\(b+\frac{a}{2}-\frac{a^2+ab+b^2}{a+b}=\frac{1}{2}\left(ab-a^2\right)\le0\Rightarrow b+\frac{a}{2}\le\frac{a^2+ab+b^2}{a+b}\)
Tương tự: \(b+\frac{c}{2}\ge\frac{b^2+bc+c^2}{b+c}\ge c+\frac{b}{2};a+\frac{c}{2}\ge\frac{a^2+ac+c^2}{a+c}\ge c+\frac{a}{2}\)
Lại có:+) \(\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\)
\(=\left(a-b\right)\frac{a^2+ab+b^2}{a+b}+\left(b-c\right)\frac{b^2+bc+c^2}{b+c}-\left(a-c\right)\frac{a^2+ac+c^2}{a+c}\)
\(\ge\left(a-b\right)\left(b+\frac{a}{2}\right)+\left(b-c\right)\left(c+\frac{a}{2}\right)-\left(a-c\right)\left(a+\frac{c}{2}\right)\)
\(\ge\frac{-1}{4}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left(1\right)\)
+) \(\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\)
\(=\left(a-b\right)\frac{a^2+ab+b^2}{a+b}+\left(b-c\right)\frac{b^2+bc+c^2}{b+c}-\left(a-c\right)\frac{a^2+ac+c^2}{a+c}\)
\(\le\left(a-b\right)\left(a+\frac{b}{2}\right)+\left(b-c\right)\left(b+\frac{c}{2}\right)-\left(a-c\right)\left(c+\frac{a}{2}\right)\)
\(\le\frac{1}{4}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left(2\right)\)
Từ 1,2 => đpcm
BĐT đã cho tuong duong voi:
\(\left|\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right|\le\frac{1}{4}\left[\Sigma\left(a-b\right)^2\right]\)
Theo AM-GM ta có: \(\left(ab+bc+ca\right)\le\frac{9}{8}\cdot\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a+b+c}\)
Có: \(VT\le\frac{9}{8}\left|\frac{\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}{\left(a+b+c\right)}\right|=\frac{9\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}{8\left(a+b+c\right)}\)
Cần chứng minh: \(4\left(a+b+c\right)^2\left[\Sigma\left(a-b\right)^2\right]^2\ge9\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\)
Rõ ràng \(\Sigma\left(a-b\right)^2\ge3\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
Cần cm: \(36\left(a+b+c\right)^2\sqrt[3]{\left(a-b\right)^4\left(b-c\right)^4\left(c-a\right)^4}\ge9\sqrt[3]{\left(a-b\right)^6\left(b-c\right)^6\left(c-a\right)^6}\)
Hay \(4\left(a+b+c\right)^2\ge\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
Tiếp tục là điều hiển nhiên do \(VT\ge4\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]\)
\(=2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
\(\ge6\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\ge VP\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\\a-b=b-c=c-a\\a=b=c\end{cases}}\Leftrightarrow a=b=c.\)
Cho a, b, c là 3 số thực dương. CMR
\(\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ca}\ge9+2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ac}=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}+6\)
\(bđt\Leftrightarrow\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge3+2\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)\)
Mà: \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{4a}{b+c}+\frac{4b}{a+c}+\frac{4c}{a+b}\)
\(\Leftrightarrow2\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)\ge3\Leftrightarrow\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\ge\frac{3}{2}\)
bđt cuối đúng theo Nesbit. Dấu "=" xảy ra khi a=b=c
cho a;b;c là các số thực dương thỏa mãn \(a^2+b^2+c^2=\frac{1}{3}\)CMR:\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\ge a+b+c\)
cho a;b;c là các số thực dương thỏa mãn a+b+c=3.CMR:\(\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le\frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}\)
bài này mà giải theo SOS là hơi bị tuyệt vời nhé =)))