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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
MÌnh nghĩ đề phải là tìm GTLN chứ
Ta có: \(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}=2\)
\(\Rightarrow\frac{1}{a+b+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\ge2\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\)
Tương tự: \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}}\)
\(\frac{1}{c+a+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân lại ta có: \(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
Dấu = khi a=b=c=1/4
a)Áp dụng Bđt cô si, ta có:
\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự ta có:
\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Cộng 3 vế của bđt lại ta có:
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}=a+b+c-\frac{ab+bc+ac}{2}\ge\frac{3}{2}\)
dấu = khi a=b=c=1
Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)
\(\ge\left(1+3a\right)\left(1-\frac{b^2}{2b}\right)=\left(1+3a\right)\left(1-\frac{b}{2}\right)\)
\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)
Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)
Đẳng thức xảy ra khi a = b = c = 1
\(2x+8y+21z\leq 12xyz\Rightarrow 3z\geq \frac{2x+8y}{4xy-7}\Rightarrow P\geq x+2y+\frac{2x+8y}{4xy-7}=x+\frac{11}{2x}+\frac{1}{2x}\left [ (4xy-7)+\frac{4x^{2}+28}{4xy-7} \right ]\geq x+\frac{11}{2x}+\frac{1}{x}\sqrt{4x^{2}+28}=x+\frac{11}{2x}+\frac{3}{2}\sqrt{\left ( 1+\frac{7}{9} \right )\left ( 1+\frac{7}{x^{2}} \right )}\geq x+\frac{11}{2x}+\frac{3}{2}\left ( 1+\frac{7}{3x} \right )=x+\frac{9}{x}+\frac{3}{2}\geq 6+\frac{3}{2}=\frac{15}{2}\)