ta có:a^4+b^4>=2(ab)^2
c^4+b^4>=2(cd)^2
cộng hai vế lai ta có:
a^4+b^4+c^4+d^4>=2[(ab)^2+(cd)^2]
>=4abcd
Để a^4+b^4+c^4+d^4=4abcd thì :
+) a^4+b^4=2(ab)^2
<->(a^2-b^2)^2=0-->a^2=b^2-->a=b(1)
+)c^4+b^4=2(cd)^2
<->(c^2-d^2)^2=0-->c^2-d^2=0-->c=d(2)
+)a^4+b^4+c^4+d^4=4abcd
<->a^4+c^4=2*(ac)^2
<->(a^2-c^2)^2=0-->a^2=c^2-->a=c(3)
từ (1)(2)(3)-->a=b=c=d(ĐPCM)
Cho \(a,b,c,d\in[0;1]\)
CMR: \(\frac{a}{bc+cd+db+1}+\frac{b}{cd+da+ac+1}+\frac{c}{da+ab+bd+1}+\frac{d}{ab+bc+ca+1}\le\frac{3}{4}+\frac{1}{4abcd}\)
chứng minh các bất đẳng thức sau:
a) \(\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\ge3abc,\left(\forall a,b,c>0\right)\)
b) \(\left(\frac{a+b+c+d}{4}\right)^4\ge abcd,\left(\forall a,b,c,d\ge0\right)\)
c) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c},\left(\forall a,b,c>0\right)\)
d) \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6,\left(\forall a,b,c>0\right)\)
CHo a b c d là các số thực . Chứng minh các bất đẳng thức :
a, \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)
b, \(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)
c, \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2^{ }}\right)^3\) với a,b >0
d, \(a^4+b^4\ge a^3b+ab^3\)
e, \(\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a^2+b^2\right)\) với ab>0
f, \(a^4+b^4\le\frac{a^6}{b^2}+\frac{b^6}{a^2}\) với a,b \(\ne\) 0
Cho a, b, c, d là những số dương.
Chứng minh rằng :
\(\dfrac{a+b+c+d}{4}\ge\sqrt[4]{abcd}\)
Cho a, b, c, d > 0. Chứng minh rằng:
1.
\(\dfrac{a}{\sqrt{a^2+8bc}}\)+ \(\dfrac{b}{\sqrt{b^2+8ac}}\)+ \(\dfrac{c}{\sqrt{c^2+8ab}}\) ≥ 1
2.
\(\dfrac{a}{b+2c+3d}\)+\(\dfrac{b}{c+2d+3a}\)+\(\dfrac{c}{d+2a+3b}\)+ \(\dfrac{d}{a+2b+3c}\) ≥ \(\dfrac{2}{3}\)
3.
\(\dfrac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}\) + \(\dfrac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}\) + \(\dfrac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}\) + \(\dfrac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\) ≥ \(\dfrac{a+b+c+d}{4}\)
Bất đẳng thức BuNyaKovSky ( BCS )
Cho a,b,c dương tm \(a+b\le c\). Tìm GTNN của \(P=\left(a^4+b^4+c^4\right)\left(\dfrac{1}{4a^4}+\dfrac{1}{4b^4}+\dfrac{1}{c^4}\right)\)
a) \(x^4+y^4\ge xy\left(x^2+y^2\right)\)với mọi x,y b) cho a,b,c>0 thoả mãn abc=1 tìm GTLN : A = \(\frac{a}{b^4+c^4+a}+\frac{b}{c^4+a^4+b}+\frac{c}{a^4+b^4+c}\)
Cho a,b,c \(\ge0\). CMR:
\(\dfrac{a^3b}{a^4+a^2b^2+b^4}+\dfrac{b^3c}{b^4+b^2c^2+c^4}+\dfrac{c^3a}{c^4+c^2a^2+a^4}\le1\)
Cho 4 số thực a, b, c, d thỏa mãn a2+b2=1 và c+d=3.
Chứng minh rằng ac+bd+cd <= \(\dfrac{9+6\sqrt{2}}{4}\)
