\(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
Cho a,b,c∈R.CM bđt \(a^2+b^2+c^2\ge ab+bc+ca\) (1). Áp dụng cm các bđt sau:
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
b)\(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)
c)\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d)\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
e)\(\frac{a+b+c}{3}\ge\sqrt{\frac{ab+bc+ca}{3}}vớia,b,c>0\)
f)\(a^4+b^4+c^4\ge abc\) nếu a+b+c=1
Cho a,b,c>0. CM các bđt sau:
a)\(\left(a^3+b^3+c^3\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)^2\)
b)\(3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
c)\(9\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)
Cho a,b,c là độ dài 3 cạnh của 1 tam giác cm:
a)\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
b)\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
c)\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\)
d)\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a+b\right)^2>a^3+b^3+c^3\)
\(\sqrt[3]{7x-8}+1\ge\left(\sqrt{2x-1}-1\right)^2\)
Cho a,b,c,d∈R.CMR a2+b2≥2ab(1) Áp dụng cm các bđt sau:
a)\(a^4+b^4+c^4+d^4\ge4abcd\)
b)\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)
c) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
Cho a,b,c>0 chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) (1). Áp dụng chứng minh các BĐT sau:
a) \(\left(a^2+b^2+c^2\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
b) Cho x,y,z>0 tm x+y+z=1. Tìm GTLN của bt \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(\left(\dfrac{a+1}{b}\right)\left(\dfrac{b+1}{c}\right)\left(\dfrac{c+1}{a}\right)\ge\)8
Giải phương trình
a, \(\sqrt[3]{x^2-1}+3=\sqrt{x^3-2}.\)
b, \(x^2-x-2\sqrt{1+16x}=2\)
c, \(\left(x-3\right)\left(x+1\right)+3\left(x-2\right).\sqrt{\frac{x+1}{x-3}=4}\)
d, \(\sqrt{\frac{x+1}{x-1}}-\sqrt{\frac{x-1}{x+1}}=\frac{3}{2}\)
e, \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=2}\)
Giải dùm với 1 câu cũng được cảm ơn tik nhiệt tình
