Cho \(a,b,c>0\) \(a+b+c=3\) . Tìm GTNN của \(A=\frac{a^3}{b^3+8}+\frac{b^3}{c^3+8}+\frac{c^3}{a^3+8}\)
Tìm GTNN của biểu thức P=\(\frac{a^3+8}{a^3\left(b+c\right)}+\frac{b^3+8}{b^3\left(c+a\right)}+\frac{c^3+8}{c^3\left(a+b\right)}\) với a, b, c là các số thực dương thỏa mãn abc=1.
cho a,b,c>0. chứng minh: \(\frac{a^8+b^8+c^8}{a^3+b^3+c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
hình như dấu + dưới mẫu là nhân mới đúng
cho số thực a,b,c>0. CMR
\(\frac{8}{\left(a+b\right)^2+4abc}+\frac{8}{\left(b+c\right)^2+4abc}+\frac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}\)
cho a,b,c>0. CMR
\(\frac{a^8}{b^3}+\frac{b^8}{c^3}+\frac{c^8}{a^3}\ge a^5+b^5+c^5\)
áp dụng bđt cô si ta có:
\(\frac{a^8}{b^3}+a^2b^3\ge2a^5;\frac{b^8}{c^3}+b^2c^3\ge2b^5;\frac{c^8}{a^3}+c^2a^3\ge2c^5\)
\(\Rightarrow\frac{a^8}{b^3}+\frac{b^8}{c^3}+\frac{c^8}{a^3}\ge2\left(a^5+b^5+c^5\right)-\left(a^2b^3+b^2c^3+c^2a^3\right)\)
áp dụng bđt cô si ta có:
\(a^5+a^5+b^5+b^5+b^5\ge5\sqrt[5]{a^5.a^5.b^5.b^5.b^5}=5a^2b^3\)
\(b^5+b^5+c^5+c^5+c^5\ge5\sqrt[5]{b^5.b^5.c^5.c^5.c^5}=5b^2c^3\)
\(c^5+c^5+a^5+a^5+a^5\ge5\sqrt[5]{c^5.c^5.a^5.a^5.a^5}=5c^2a^3\)
\(\Rightarrow5\left(a^5+b^5+c^5\right)\ge5\left(a^2b^3+b^2c^3+c^2a^3\right)\Rightarrow a^5+b^5+c^5\ge a^2b^3+b^2c^3+c^2a^3\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-\left(a^2b^3+b^2c^3+c^2a^3\right)\ge a^5+b^5+c^5\)
\(\frac{a^8}{b^3}+\frac{b^8}{c^3}+\frac{c^8}{a^3}\ge a^5+b^5+c^5\left(Q.E.D\right)\)
dấu = xảy ra khi a=b=c
Cho a, b, c > 0; abc ≥ 8. Tìm GTNN của:
\(P=\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\)
1.Cho a+b+c+d ≠0 và \(\frac{a}{b+c+d}\)=\(\frac{b}{a+c+d}\)=\(\frac{c}{a+b+d}\)=\(\frac{d}{a+b+c}\)
Tính giá trị của A=\(\frac{a+b}{c+d} \)+\(\frac{b+c}{a+d}\)+\(\frac{c+d}{a+b}\)+\(\frac{d+a}{b+c}\)
2.Tìm x,y,z biết :
a)\(\dfrac{x^3}{8}\)=\(\dfrac{y^3}{64}\)=\(\dfrac{z^3}{216}\)và \(x^2\)+\(y^2\)+\(z^2\)=14
b)\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
Cho a,b,c>0 và a+b+c=3.Tìm GTNN của \(\frac{a^3}{2b+c}+\frac{b^3}{2c+a}+\frac{c^3}{2a+b}\)
nếu ai trả lời trc tao , thì thằng đó tự đăng tự tl
\(\frac{a^3}{2b+C}+\frac{\left(2b+c\right)}{9}+\frac{1}{3}\ge3\sqrt[3]{\frac{a^3}{27}}=a.\)
\(\frac{b^3}{2c+A}+\frac{\left(2c+a\right)}{9}+\frac{1}{3}\ge b\)
\(\frac{c^3}{2a+b}+\frac{\left(2a+b\right)}{9}+\frac{1}{3}\ge c\)
\(VT+\frac{1}{3}\left(a+b+c\right)+\frac{4}{3}\ge3\)
\(VT+\frac{7}{3}\ge3\Leftrightarrow VT\ge1\)
Min của Vt là 1 , dấu = " khi x=y=z=1
cho a+b+c=3
CMR \(\frac{a^3}{b^3+8}+\frac{b^3}{c^3+8}+\frac{c^3}{a^3+8}>=\frac{1}{3}\)
Tìm min,max của P=xyz biết A= \(\frac{8-x^2}{16+x^4}+\frac{8-y^2}{16+y^4}+\frac{8-z^2}{16+z^4}\ge0.\)
Cho a;b;c >0 thỏa mã \(a+b+c\le3\)Tìm min P \(=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)