Cho a,b,c >3 tm a+b+c=3 Tìm \(P=\frac{a}{3+b-a}+\frac{b}{3+c-b}+\frac{c}{3+a-c}\)
Cho a,b,c nguyên dương tm a+b+c=1 tm\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{3}{2}\)
Cm a=b=c
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm a^2+b^2+c^2 bé hơn hoặc bằng abc. Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm a+b+c<=3. Cmr \(\frac{ab}{\sqrt{3+c}}+\frac{bc}{\sqrt{3+a}}+\frac{ca}{\sqrt{3+b}}\le\frac{3}{2}\)
4) Cho a,b,c>0 tm a+b+c=2. Cmr \(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)
5) Cho a,b,c>0. Cmr \(\sqrt{\frac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\frac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\frac{c^3}{5c^2+\left(a+b\right)^2}}\le\sqrt{\frac{a+b+c}{3}}\)
6) Cho a,b,c>0. Cmr \(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\frac{1}{3}\)
Giúp mình với nhé các bạn
cho 3 so duong a,b,c tm a+b+c=6
cmr\(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\ge2\)
Mình chỉ làm sơ sơ, có gì bạn sửa lại
Ta có: \(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\)
Đặt a ; b và c = 2 .
Thế số vào biểu thức ta có:
\(\frac{2}{\sqrt{2^3+1}}+\frac{2}{\sqrt{2^3+1}}+\frac{2}{\sqrt{2^3+1}}\)
\(\Leftrightarrow\frac{2}{\left(2^3+1\right)^2}+\frac{2}{\left(2^3+1\right)^2}+\frac{2}{\left(2^3+1\right)^2}\)
\(\Leftrightarrow\frac{2}{\left(2^3+1\right)^2}.3\Leftrightarrow\frac{2}{\left(8+1\right)^2}.3\Leftrightarrow\frac{2}{9^2}\ge2\)
Ta có ĐPCM
cho a,b,c, là 3 số dương tm đk \(a+b+c=1\)
cmr \(\frac{a^3}{\left(b+c\right)^2}+\frac{b^3}{\left(c+a\right)^2}+\frac{c^3}{\left(a+b\right)^2}\ge\frac{1}{4}\)
\(BDT\Leftrightarrow\frac{a^3}{\left(1-a\right)^2}+\frac{b^3}{\left(1-b\right)^2}+\frac{c^3}{\left(1-c\right)^2}\ge\frac{1}{4}\)
Ta có BĐT phụ: \(\frac{a^3}{\left(1-a\right)^2}\ge a-\frac{1}{4}\)
\(\Leftrightarrow\frac{\left(3a-1\right)^2}{4\left(a-1\right)^2}\ge0\forall0< a\le\frac{1}{3}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\frac{b^3}{\left(1-b\right)^2}\ge b-\frac{1}{4};\frac{c^3}{\left(1-c\right)^2}\ge c-\frac{1}{4}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\left(a+b+c\right)-\frac{1}{4}\cdot3=1-\frac{3}{4}=\frac{1}{4}=VP\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
Áp dụng BĐT cô si ta có:
\(\frac{a^3}{\left(b+c\right)^2}+\frac{1a}{4}\ge\frac{a^2}{b+c}\)\(,\frac{b^3}{\left(c+a\right)^2}+\frac{1b}{4}\ge\frac{b^2}{a+c},\frac{c^3}{\left(a+b\right)^2}+\frac{1c}{4}\ge\frac{c^2}{a+b}\)
Cộng lại ta có
\(VT\ge\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}-\frac{1}{4}\left(a+b+c\right)\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}-\frac{1}{4}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\left(đpcm\right)\)
Dấu =tự tìm Ok
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm \(a^2+b^2+c^2\le abc\).Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm \(\sqrt{a}+\sqrt{b}+\sqrt{c}=1\).Cmr \(\sqrt{\frac{ab}{a+b+2c}}+\sqrt{\frac{bc}{b+c+2a}}+\sqrt{\frac{ca}{c+a+2b}}\le\frac{1}{2}\)
Giúp mình mới nhé các bạn. Mình đang cần gấp
Cho a,b,c dương tm ab+bc+ca=3
Tìm min\(A=\frac{19a+3}{b^2+1}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)
Ta có: \(\frac{19a+3}{b^2+1}=\left(19a+3\right).\frac{1}{b^2+1}=\left(19a+3\right)\left(1-\frac{b^2}{b^2+1}\right)\)
\(\ge\left(19a+3\right)\left(1-\frac{b^2}{2b}\right)=\left(19a+3\right)\left(1-\frac{b}{2}\right)\)
\(=19a+3-\frac{19ab}{2}-\frac{3b}{2}\)(1)
Hoàn toàn tương tự, ta có: \(\frac{19b+3}{c^2+1}\ge19b+3-\frac{19bc}{2}-\frac{3c}{2}\)(2); \(\frac{19c+3}{a^2+1}\ge19c+3-\frac{19ca}{2}-\frac{3a}{2}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(A=\frac{19a+3}{b^2+1}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)\(\ge19\left(a+b+c\right)-\frac{3\left(a+b+c\right)}{2}-\frac{19\left(ab+bc+ca\right)}{2}+9\)
\(=\frac{35\left(a+b+c\right)}{2}-\frac{19\left(ab+bc+ca\right)}{2}+9\)
\(\ge\frac{35.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{19.3}{2}+9=\frac{105}{2}-\frac{57}{2}+9=33\)
Đẳng thức xảy ra khi a = b = c = 1.
Cho 3 số a,b,c >0 tm: a+b+c=1 Tìm Max
P=\(\frac{a}{9a^3+3b^2+c}+\frac{b}{9b^3+3c^2+a}+ \frac{c}{9c^3+3a^2+b}\)
Cauchy-SChwarz:
\(\left(9a^3+3b^2+c\right)\left(\dfrac{1}{9a}+\dfrac{1}{3}+c\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{a}{\left(9a^3+3b^2+c\right)}\le\dfrac{a\left(\dfrac{1}{9a}+\dfrac{1}{3}+c\right)}{\left(a+b+c\right)^2}=\dfrac{\dfrac{1}{9}+\dfrac{a}{3}+ac}{\left(a+b+c\right)^2}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(P\le\dfrac{1}{9}\cdot3+\dfrac{a+b+c}{3}+ab+bc+ca\)
\(\le\dfrac{1}{9}\cdot3+\dfrac{a+b+c}{3}+\dfrac{\left(a+b+c\right)^2}{3}=1\)
Dấu "=" \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
cho a.b.c khác 0 tm
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}\)
tính gtrị biểu thức (a-b)3+(b-c)3+(c-a)3
\(\sqrt{\frac{19}{a+b-c}}+\sqrt{\frac{5}{b+c-a}}+\sqrt{\frac{79}{a+c-b}}\in N\ne1\)
tìm các số nguyên dương a,b,c tm a,b,c là 3 cạnh 1 tam giác