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Áp dụng BĐT AM - GM:
\(\frac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\) \(\Rightarrow abc\le\frac{1}{8}\)
\(1+1+1+\frac{1}{2a}+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2b}\ge7\sqrt[7]{\frac{1}{16a^2b^2}}\)
\(\Leftrightarrow3+\frac{1}{a}+\frac{1}{b}\ge7\sqrt[7]{\frac{1}{16a^2b^2}}\)
Tương tự ta CM được:
\(3+\frac{1}{b}+\frac{1}{c}\ge7\sqrt[7]{\frac{1}{16b^2c^2}}\)
\(3+\frac{1}{c}+\frac{1}{a}\ge\ge7\sqrt[7]{\frac{1}{16c^2a^2}}\)
Nhân vế theo vế 3 bất đẳng thức trên:
\(S\ge343\sqrt[7]{\frac{1}{4096a^4b^4c^4}}\ge343\sqrt[7]{\frac{1}{4096.\frac{1}{8^4}}}=343\)
\(\Rightarrow Min_S=343\Leftrightarrow a=b=c=\frac{1}{2}\)
a2(b+c)2+5bc+b2(a+c)2+5ac≥4a29(b+c)2+4b29(a+c)2=49(a2(1−a)2+b2(1−b)2)(vì a+b+c=1)
a2(1−a)2−9a−24=(2−x)(3x−1)24(1−a)2≥0(vì )<a<1)
⇒a2(1−a)2≥9a−24
tương tự: b2(1−b)2≥9b−24
⇒P⩾49(9a−24+9b−24)−3(a+b)24=(a+b)−94−3(a+b)24.
đặt t=a+b(0<t<1)⇒P≥F(t)=−3t24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13
để biểu thức cho đơn giản , ta đặt x=a+1,y=b+1,z=c+1(x,y,z>0)
thì giả thiết thành \(\frac{1}{x+1}+\frac{3}{y+3}\le\frac{z}{z+2}\) .Tìm min xyz
Áp dụng bất đẳng thức cauchy:\(\frac{z}{z+2}\ge\frac{1}{x+1}+\frac{3}{y+3}\ge2\sqrt{\frac{3}{\left(x+1\right)\left(y+3\right)}}\)(1)
từ giả thiết :\(\frac{1}{x+1}\le\frac{z}{z+2}-\frac{3}{y+3}\Leftrightarrow1-\frac{1}{x+1}\ge1-\frac{z}{z+2}+\frac{3}{y+3}\)
\(\Leftrightarrow\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\)
Áp dụng bất đẳng thức cauchy 1 lần nữa: \(\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\ge2\sqrt{\frac{6}{\left(z+2\right)\left(y+3\right)}}\)(2)
tương tự ta cũng có: \(\frac{y}{y+3}\ge2\sqrt{\frac{2}{\left(z+2\right)\left(x+1\right)}}\)(3),
cả 2 vế các bất đẳng thức (1),(2)và (3) đều dương, nhân vế với vế:
\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{8.6}{\left(x+1\right)\left(z+2\right)\left(y+3\right)}\)
\(\Leftrightarrow xyz\ge48\)
Dấu = xảy ra khi x=2,y=6,z=4 hay a=1,b=5,z=3
We have:
\(M=1-\frac{1}{3}\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\)
Consider:
\(\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\ge\frac{3}{2}\)
\(VT\ge\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\)
Prove:
\(\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\ge\frac{3}{2}\)
\(\Leftrightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge2\left(a^2+b^2+c^2\right)+27\)
Consider:
\(\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\Sigma_{cyc}a^2+\Sigma_{cyc}ab\)
\(\Rightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab\)
Now we need to prove:
\(4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab=2\Sigma_{cyc}a^2+27\)
\(\Leftrightarrow2\left(a+b+c\right)^2=27\) (not fail)
\(\Rightarrow M\le\frac{1}{2}\)
Sign '=' happen when \(a=b=c=\sqrt{\frac{3}{2}}\)
Áp dụng Bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có:
\(P\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
Lại có:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\)
\(\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=9\)
Mặt khác \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
\(\Rightarrow\frac{1}{ab+bc+ca}\ge3\)\(\Rightarrow P_{Min}=30\)
Dấu = khi \(a=b=c=\frac{1}{3}\)
Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :
\(M\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)
Do đó : \(M\ge\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có : \(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\left(\frac{1}{a}\right)^2}{a\left(b+c\right)}\)
Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)
Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Áp dụng BĐT Svacxo ta có :
\(M=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^2\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\) \(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)
\(\frac{c+1}{c+3}\ge\frac{1}{a+2}+\frac{3}{b+4}\ge2\sqrt[]{\frac{3}{\left(a+2\right)\left(b+4\right)}}\) (1)
\(\frac{1}{a+2}+\frac{3}{b+4}\le\frac{c+3-2}{c+3}=1-\frac{2}{c+3}\Rightarrow1-\frac{1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\)
\(\Rightarrow\frac{a+1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\) (2)
\(\frac{1}{a+2}+\frac{3}{b+4}\le1-\frac{2}{c+3}\Rightarrow1-\frac{3}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\)
\(\Rightarrow\frac{b+1}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\) (3)
Nhân vế với vế (1);(2);(3):
\(\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\ge8\sqrt{\frac{36}{\left(a+2\right)^2\left(b+4\right)^2\left(c+3\right)^2}}=\frac{48}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\)
\(\Rightarrow Q\ge48\Rightarrow Q_{min}=48\) khi \(\left(a;b;c\right)=\left(1;5;3\right)\)