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a)Áp dụng BĐT Cauchy-Schwarz dạng Engel:
\(VT=\left(\frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\ge\frac{9\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}\ge\frac{9\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{\left(a+b+c\right)^2}=\left(a+b+c\right)^2\)
Đẳng thức xảy ra khi \(a=b=c\)
b) \(VT-VP=\left(a+b\right)\left(a-b\right)^2+\left(b+c\right)\left(b-c\right)^2+\left(c+a\right)\left(c-a\right)^2\ge0\)
Đẳng thức xảy ra khi \(a=b=c\)
c) Theo câu b và BĐT Cauchy-Schwarz:
\(\Rightarrow3.3\left(a^3+b^3+c^3\right)\ge3\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(\ge3\left(a+b+c\right)\left[\frac{\left(a+b+c\right)^2}{3}\right]=\left(a+b+c\right)^3\)
Đẳng thức xảy ra khi \(a=b=c\)
Ta có: \(a+b+c\ge3\sqrt[3]{abc}\)
\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)(1)
Ta có: \(\left(a-b\right)^3+\left(b-c\right)^2+\left(c-a\right)^3\)
\(=\left(a-b\right)^3+3\left(a-b\right)^2\left(b-c\right)+3\left(a-b\right)\left(b-c\right)^2+\left(b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)^2\left(b-c\right)-3\left(a-b\right)\left(b-c\right)^2\)
\(=\left(a-b+b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Ta có: \(a-b+b-c+c-a\ge3\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow9abc\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)(2)
Từ (1), (2) ta có: \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:\(\left(a^2+bc\right)\left(b+c\right)=b\left(a^2+c^2\right)+c\left(a^2+b^2\right)\)
\(\Rightarrow\sqrt{\frac{\left(a^2+bc\right)\left(b+c\right)}{a\left(b^2+c^2\right)}}=\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Tương tự\(\Rightarrow\)VT=\(\Sigma\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Đặt \(x=a\left(b^2+c^2\right)\);\(y=b\left(a^2+c^2\right)\);\(z=c\left(b^2+a^2\right)\)
VT=\(\sqrt{\frac{x+y}{z}}+\sqrt{\frac{y+z}{x}}+\sqrt{\frac{x+z}{y}}\ge3\sqrt[6]{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}}\ge3\sqrt{2}\)(BĐT Cô-si)
Dấu''='' xra\(\Leftrightarrow\)a=b=c
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{8}+\frac{c+1}{8}\ge\frac{3}{4}a\)\(\Leftrightarrow\)\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{3}{4}a-\frac{1}{8}b-\frac{1}{8}-\frac{1}{4}\)
\(\Sigma\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\) :)
a)đpcm<=>(a2+3)2>4(a2+2)<=>(a2+1)2>0(lđ)
b)đpcm<=>\(a^4+b^4\ge ab\left(a^2+b^2\right)\)
Theo AM-GM\(\left\{{}\begin{matrix}a^4+b^4+b^4+b^4\ge4a^3b\\b^4+a^4+a^4+a^4\ge4b^3a\end{matrix}\right.\)
=>đpcm. Dấu bằng xảy ra khi a=b
c)AM-GM:\(VT\ge256\left|abcd\right|\ge256abcd\)
Dấu bằng xảy ra khi hai số bằng 2, hai số còn lại bằng -2 hoặc cả 4 số bằng 2 hoặc cả 4 số bằng -2
Bất đẳng thức cần chứng minh viết lại thành:
\(\frac{\left(a+b\right)^3}{\left(b+c\right)^3}+\frac{\left(a+c\right)^3}{\left(b+c\right)^3}+\frac{3\left(a+b\right)\left(a+c\right)\left(b+c\right)}{\left(b+c\right)^3}\le5\)
Đặt: \(x=\frac{a+b}{b+c};y=\frac{a+c}{b+c}\), bất đẳng thức chứng minh trở thành:
\(x^3+y^3+3xy\le5\)
Ta có:
\(xy=\frac{a+b}{b+c}+\frac{a+c}{b+c}=\frac{\left(a+b\right)\left(a+c\right)}{\left(b+c\right)^2}\)
\(=\frac{a\left(a+b+c\right)+bc}{\left(b+c\right)^2}=\frac{2a\left(a+b+c\right)-2bc}{\left(b+c\right)^2}\)
Vậy ta được: \(xy+1=\frac{\left(a+b\right)^2+\left(a+c\right)^2}{\left(b+c\right)^2}=x^2+y^2\)
\(x^3+y^3=x+y\)nên \(x^3+y^3+3xy\le5\Leftrightarrow x+y+3xy\le5\)
Mà ta có: \(\left(x+y\right)^2\le2\left(x^2+y^2\right)=\frac{xy+1}{2}\le\frac{1}{2}+\frac{\left(x+y\right)^2}{8}\)
\(\Rightarrow x+y\le2\Rightarrow xy\le1\)
Do đó ta được: \(x+y+3xy\le5\). Vậy bài toán đã được chứng minh.