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\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cauchy-schwarz:
\(VT=\frac{c^2}{ac^2+bc^2}+\frac{a^2}{a^2b+a^2c}+\frac{b^2}{b^2c+b^2a}+\frac{\sqrt[3]{a^2b^2c^2}}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng BĐT AM-GM ta có \(\frac{1^2}{a\left(a+b\right)}+\frac{1^2}{b\left(b+c\right)}+\frac{1^2}{c\left(c+a\right)}\ge\)
\(\ge\frac{\left(1+1+1\right)^2}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}=\frac{9}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}\ge\)
\(\ge\frac{9}{3.\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
\(VT=\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}+\Sigma\frac{a^2}{a^2\left(b+c\right)}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\Sigma a^2\left(b+c\right)+2abc}=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
a )
Áp dụng BĐT Bunhiacopxki ta có :
\(\left(b^2+\left(c+a\right)^2\right)\left(1+\right)\ge\left(b+2\left(a+c\right)\right)^2\)
\(\Rightarrow\sqrt{\frac{a^2}{b^2+\left(c+a\right)^2}}\le\sqrt{5}.\frac{a}{b+2c+2a}\)
\(\Rightarrow VT\le\sqrt{5}.\left(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\right)\)
Cần chứng minh : \(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\le\frac{3}{5}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{a}{b+2c+2a}\right)+\left(\frac{1}{2}-\frac{b}{c+2a+2b}\right)+\left(\frac{1}{2}-\frac{c}{a+2b+2c}\right)\ge\frac{9}{10}\)
\(\Leftrightarrow\frac{b+2c}{b+2c+2a}+\frac{c+2a}{c+2a+2b}+\frac{a+2b}{a+2b+2c}\ge\frac{9}{5}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức ở vế trái :
\(\Rightarrow VT\ge\frac{\left(b+2c+c+2a+a+2b\right)^2}{\left(b+2c\right)^2+2a\left(b+2c\right)+\left(c+2a\right)^2+2b\left(c+2a\right)+\left(a+2b\right)^2+2c\left(a+2b\right)}\)
\(=\frac{9\left(a+b+c\right)^2}{5\left(a+b+b\right)^2}=\frac{9}{5}\left(đpcm\right)\)
Dấu " = '" xảy ra khi a=b=c
b ) Ta có abc =1
Ta chứng minh :
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}=1\)
VT \(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ac}\)
\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=1\left(đpcm\right)\)
Ta có : \(\left(1+a\right)^2+b^2+5=\left(a^2+b^2\right)+2a+6\ge2ab+2a+6\)
\(\Rightarrow\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}=\frac{2ab+2a+6}{ab+a+4}=2-\frac{2}{ab+a+4}\)
Mà \(\frac{1}{ab+a+4}=\frac{1}{ab+a+1+3}\le\frac{1}{4}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)\) ( do \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}\ge2-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)=\frac{11}{6}-\frac{1}{2}.\frac{1}{ab+a+1}\)
Khi đó :
\(P\ge\frac{11}{2}-\frac{1}{2}.\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\right)=\frac{11}{2}-\frac{1}{2}.1=5\)
\(P_{Min}=5\) khi \(a=b=c=1\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(c+a\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\)
Áp dụng BĐT Bun :
\(\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(a+c\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{c^2\left(a+b\right)+a^2\left(b+c\right)+b^2\left(a+c\right)+2abc}=...\)
Dấu ''='' xảy ra khi a = b =c
Áp dụng Bất Đẳng Thức \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\forall x;y;z\inℝ\)ta có
\(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc>0\Rightarrow ab+bc+ca\ge3\sqrt{abc}\)
Ta có \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\forall a;b;c>0\)
Thật vậy \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)
\(\ge1+3\sqrt[3]{abc}+3\sqrt[3]{\left(abc\right)^2}+abc=\left(1+\sqrt[3]{abc}\right)^3\)
Khi đó \(P\le\frac{2}{3\left(1+\sqrt{abc}\right)}+\frac{\sqrt[3]{abc}}{1+\sqrt[3]{abc}}+\frac{\sqrt{abc}}{6}\)
Đặt \(\sqrt[6]{abc}=t\Rightarrow\sqrt[3]{abc}=t^2,\sqrt{abc}=t^3\)
Vì a,b,c>0 nên 0<abc\(\le\left(\frac{a+b+c}{3}\right)^2=1\Rightarrow0< t\le1\)
Xét hàm số \(f\left(t\right)=\frac{2}{3\left(1+t^3\right)}+\frac{t^2}{1+t^2}+\frac{1}{6}t^3;t\in(0;1]\)
\(\Rightarrow f'\left(t\right)=\frac{2t\left(t-1\right)\left(t^5-1\right)}{\left(1+t^3\right)^2\left(1+t^2\right)^2}+\frac{1}{2}t^2>0\forall t\in(0;1]\)
Do hàm số đồng biến trên (0;1] nên \(f\left(t\right)< f\left(1\right)\Rightarrow P\le1\)
\(\Rightarrow\frac{2}{3+ab+bc+ca}+\frac{\sqrt{abc}}{6}+\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\le1\)
Dấu "=" xảy ra khi a=b=c=1
Áp dụng BĐT Cô-si cho 3 số dương, ta có :
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)
Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)
hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)
\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)
\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Nhân từng vế 2 bất đẳng thức trên, ta được đpcm
Dấu "=" xảy ra khi a = b = c
Vậy ...
2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có :
\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)
Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)
Tương tự : ....
\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)
\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)
Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1