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a) Ta có: \(a^2-1\le0;b^2-1\le0;c^2-1\le0\)
\(\Rightarrow\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\le0\)
\(a^2+b^2+c^2\le1+a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2\le1+a^2b^2+b^2c^2+c^2a^2\) ( vì \(abc\ge0\) )
Có \(b-1\le0\Rightarrow a^2b\sqrt{b}\left(b-1\right)\le0\Rightarrow a^2b^2\le a^2b\sqrt{b}\)
Tương tự: \(\hept{\begin{cases}b^2c^2\le b^2c\sqrt{c}\\c^2a^2\le c^2a\sqrt{a}\end{cases}\Rightarrow dpcm}\)
\(1,VT=2\left(a^3+b^3+c^3\right)+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Ta có \(a^3+b^3\ge ab\left(a+b\right)\)
\(b^3+c^3\ge bc\left(b+c\right)\)
\(c^3+a^3\ge ca\left(c+a\right)\)
Cộng từng vế các bđt trên ta được
\(VT\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Bây giờ ta cm:
\(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
Bất đẳng thức trên luôn đúng
Vậy bđt được chứng minh
Dấu "=" xảy ra khi a=b=c
Áp dụng bất đẳng thức Cauchy–Schwarz dạng Engel ta có :
\(VT\ge\frac{\left(2b+3c+2c+3a+2a+3b\right)^2}{a+b+c}\)
\(=\frac{\left(5a+5b+5c\right)^2}{a+b+c}=\frac{\left[5\left(a+b+c\right)\right]^2}{a+b+c}\)
\(=\frac{25\left(a+b+c\right)^2}{a+b+c}=25\left(a+b+c\right)=VP\)
=> đpcm
Đẳng thức xảy ra <=> a = b = c
Ta có:
a)
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(=\left[\left(a+b+c\right)^2-2ab-2ac-2bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left[ab+ac+bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left(ab\right)^2+4\left(ac\right)^2+4\left(bc\right)^2-8abc\left(a+b+c\right)-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
b)\(=2\left(ab+bc+ac\right)^2-4\left(abbc+abca+bcca\right)\)
\(=2\left(ab+bc+ac\right)^2-4abc\left(a+b+c\right)=2\left(ab+bc+ac\right)^2\)
c) \(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=\frac{a^4+b^4+c^4+a^4+b^4+c^4}{2}\)
\(=a^4+b^4+c^4\)
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a.\left(a+b+c\right)+bc\right]\left[b.\left(a+b+c\right)+ac\right]\left[c.\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ba+b^2+bc+ac\right)\left(ca+cb+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ba+b^2\right)+\left(bc+ac\right)\right]\left[\left(ca+c^2\right)\left(cb+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(b+a\right)\right]\left[c\left(a+c\right)b\left(b+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ac\right]\left[c\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)\left(ac+bc+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ab+b^2\right)+\left(bc+ac\right)\right]\left[\left(ac+c^2\right)+\left(bc+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
\(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(2c-b\right)^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2-2a\left(b+c\right)+\left(b+c\right)^2+4c^2-4bc+b^2=2a^2+2b^2+5c^2+2ab+2bc+2ac-2ab+b^2+2bc+b^2=2a^2+3b^2+6c^2+4bc\)
=> Đề sai