Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Ta có:
\(2\left(a^3+b^3\right)\ge\left(a+b\right)\left(a^2+b^2\right)\)
\(\Rightarrow2\left(a+b\right)\left(a^2-ab+b^2\right)\ge\left(a+b\right)\left(a^2+b^2\right)\)
\(\Rightarrow2\left(a^2-ab+b^2\right)\ge a^2+b^2\)
\(\Rightarrow2a^2-2ab+2b^2\ge a^2+b^2\)
\(\Rightarrow\left(a^2+b^2-2ab\right)+a^2+b^2\ge a^2+b^2\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Cho a, b, c > 0 . CMR:
\(\frac{1}{a+b+c}\ge\frac{a^3}{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}+\frac{b^3}{\left(2b^2+c^2\right)\left(2b^2+a^2\right)}+\frac{c^3}{\left(2c^2+a^2\right)\left(2c^2+a^2\right)}\)
CMR: \(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3\) với a, b, c > 0
CMR: \(\left(a^2+b^2\right)\left(a^4+b^4\right)\) ≥ \(\left(a^3+b^3\right)^2\)
CMR: \(a\left(a+b\right)\left(a+c\right)\left(a+b+c\right)+b^2c^2\)≥ 0
1.cho x+y+z=xyz và xy+yz+zx≠3
cmr: x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)/xy+yz+zx=xyz
2.cmr nếu c^2+2(ab-ac-bc)=0và b≠c,a+b≠c thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
3. cho a,b,c thỏa mãn abc≠0 và ab+bc+ca=0
tính :P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
CMR: \(\left(a+b\right)\left(a^3+b^3\right)\) ≤ \(2\left(a^4+b^4\right)\)
Cho 3 số thực a, b, c đôi một khác nhau thỏa mãn: \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)
CMR: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)
Cho 3 số a, b, c thỏa mãn a # -b, b # -c, c # -a.
Chứng minh rằng : \(\dfrac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^2-ab}{\left(c+a\right)\left(c+b\right)}=0\)
Cho \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
Chứng minh rằng: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)
