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Ta có: \(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab}{1+b^2}\)
\(1+b^2\ge2b\) \(\Rightarrow\frac{ab^2}{1+b^2}\le\frac{ab^2}{2b}=\frac{ab}{2}\)\(\Rightarrow-\frac{ab^2}{1+b^2}\ge-\frac{ab}{2}\)
Do đó: \(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\)
Tương tự: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Suy ra \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}+\frac{ab+bc+ca}{2}\ge a+b+c\)
Mặt khác ta có: \(3\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\Rightarrow\frac{3}{a+b+c}\le1\)
\(\Rightarrow a+b+c\ge3\)
Do đó; \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}+\frac{ab+bc+ca}{2}\ge a+b+c\ge3\)(đpcm)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)
ap dung bdt am gm
\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)
\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)
tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)
\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)
tiep tuc ap dung bat cauchy-schwarz dang engel ta co
\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)
dau = xay ra \(\Leftrightarrow a=b=c=1\)
\(\frac{1}{c^2\left(a+b\right)}\ge\frac{3}{2};\frac{z^3}{x\left(y+2z\right)}\ge\frac{x+y+z}{3}\)
Chú ý đến giả thiết a + b + c = 1 ta viết được \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1-c\right)\left(1+c\right)}}=\)\(\frac{ab}{\left(a+b\right)\sqrt{1-c^2}}=\frac{ab}{\left(a+b\right)\sqrt{\left(a+b+c\right)^2-c^2}}\)\(=\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\)
Mặt khác áp dụng bất đẳng thức Cauchy ta được \(a^2+b^2+2\left(ab+bc+ca\right)\ge2ab+2\left(ab+bc+ca\right)=\)\(2\left(ab+bc\right)+2\left(ab+ca\right)\)và \(a+b\ge2\sqrt{ab}\)
Từ đó dẫn đến \(\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\le\frac{ab}{2\sqrt{ab}\sqrt{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)\(=\frac{1}{2}\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)
Mà theo bất đẳng thức quen thuộc \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta có: \(\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\le\sqrt{\frac{1}{4}\left(\frac{ab}{2\left(ab+bc\right)}+\frac{ab}{2\left(ab+ca\right)}\right)}\)
\(=\frac{1}{2\sqrt{2}}\sqrt{\frac{ab}{ab+bc}+\frac{ab}{ab+ca}}=\frac{1}{2\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)
Từ đó ta có bất đẳng thức: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)(1)
Hoàn toàn tương tự, ta có: \(\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+a\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}\)(2) ; \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)(3)
Cộng theo vế 3 bất đẳng thức (1), (2), (3), ta được: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+c\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)\(\le\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\)
Ta cần chứng minh\(\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\le\frac{3\sqrt{2}}{8}\)
Hay \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\le3\)
Áp dụng bất đẳng thức Bunhiacopxki ta được \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)
\(\le\sqrt{3\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{c+b}+\frac{a}{a+b}\right)}=3\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Sửa đề: \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)
C/m: BDT: \(a^3+b^3+abc\ge ab\left(a+b+c\right)\) (1)
That vay ta co:
\(a^3+b^3+abc-ab\left(a+b+c\right)=\left(a+b\right)\left(a-b\right)^2\ge0\) (luon dung)
Tuong tu ta co: \(b^3+c^3+abc\ge bc\left(a+b+c\right)\) (2)
\(c^3+a^3+abc\ge ca\left(a+b+c\right)\) (3)
Tu (1), (2), (3) suy ra:
\(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}\) (dpcm)