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Đề khắm vậy -_- a + b = 3 - c thì viết luôn thành a + b + c = 3 cho rồi .... bày đặt
Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\left(x;y;z>0\right)\)
\(VT=a^3+b^3+c^3+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge a^3+b^3+c^3+\frac{18}{a+b+c}\)
\(=a^3+b^3+c^3+6\)
Áp dụng bđt Cô-si cho 3 số ta đc
\(a^3+1+1\ge3\sqrt[3]{a^3.1.1}=3a\)
\(b^3+1+1\ge3b\)
\(c^3+1+1\ge3c\)
Cộng từng vế vào ta được
\(VT\ge a^3+b^3+c^3+6\ge3\left(a+b+c\right)=\left(a+b+c\right)^2\)
Lại có : \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(Phá ngoặc + chuyển vế -> tổng bình phương)
\(\Rightarrow VT\ge3\left(ab+bc+ca\right)\)(Đpcm)
Dấu "=" xảy ra <=> a = b = c = 1
Vậy ....
Áp dụng BĐT AM-GM ta có:
\(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{ab^2+b^2}{b^2+1}\ge\left(a+1\right)-\frac{ab^2+b^2}{2b}=\left(a+1\right)-\frac{ab+b}{2}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\ge a+b+c+3-\frac{a+b+c+ab+bc+ac}{2}\)
\(\ge a+b+c+3-\frac{a+b+c+\frac{\left(a+b+c\right)^2}{3}}{2}\)
\(\ge3+3-\frac{3+\frac{3^2}{3}}{2}=3\)
\("="\Leftrightarrow a=b=c=1\)
\(P=\sum\frac{a+1}{b^2+1}=\sum\left(a+1-\frac{b^2\left(a+1\right)}{b^2+1}\right)\ge\sum\left(a+1-\frac{b^2\left(a+1\right)}{2b}\right)=\sum\left(a+1-\frac{1}{2}b\left(a+1\right)\right)\)
\(\Rightarrow P\ge\frac{1}{2}\left(a+b+c\right)-\frac{1}{2}\left(ab+bc+ca\right)+3\)
\(P\ge\frac{1}{2}\left(a+b+c\right)-\frac{1}{6}\left(a+b+c\right)^2+3=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(a+1-\frac{a+1}{b^2+1}=\frac{ab^2+b^2}{b^2+1}\le\frac{ab^2+b^2}{2b}=\frac{ab}{2}+\frac{b}{2}\) vì \(b^2+1\ge2b\)
nên \(\frac{a+1}{b^2+1}\ge a+1-\frac{b}{2}-\frac{ab}{2}\) Tương tự:
Vậy ta có: \(VT\ge a+b+c+3-\frac{a+b+c}{2}-\frac{1}{2}\left(ab+bc+ca\right)\)
Vì \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{9}{3}=3\)
nên \(VT\ge3+\frac{a+b+c}{2}-\frac{1}{2}3=3+\frac{3}{2}-\frac{3}{2}=3=VP\)
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Đặt: \(P=\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\)
Ta có:
\(\frac{a+1}{b^2+1}=a-\frac{ab^2-1}{b^2+1}\ge a-\frac{ab^2-1}{2b}=a-\frac{ab}{2}+\frac{1}{2b}\)
Tương tự ta có:
\(\frac{b+1}{c^2+1}\ge b-\frac{bc}{2}+\frac{1}{2c},\frac{c+1}{a^2+1}\ge c-\frac{ca}{2}+\frac{1}{2a}\)
\(\Rightarrow P\ge a+b+c-\frac{ab+bc+ca}{2}+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3-\frac{\left(a+b+c\right)^2}{6}+\frac{1}{2}\left(\frac{\left(1+1+1\right)^2}{a+b+c}\right)\)
\(=3-\frac{9}{6}+\frac{1}{2}.\frac{9}{3}=3\)
Dấu bằng xảy ra khi a=b=c=1
\(\frac{a+1}{b^2+1}=\frac{\left(a+1\right)\left(b^2+1\right)-b^2\left(a+1\right)}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\)
\(\ge a+1-\frac{b^2\left(a+1\right)}{2b}=a+1-\frac{ab+a}{2}\)
Thiết lập các bất đẳng thức tương tự rồi cộng lại ta được:
\(LHS\ge a+b+c+3-\frac{ab+bc+ca+3}{2}\ge6-\frac{\frac{\left(a+b+c\right)^2}{3}+3}{2}=3=RHS\)