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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(a\ge b\ge c\ge1\) ta có bổ đề
\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)
Lợi dụng cái trên ta được
\(\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}+\frac{1}{1+abc}\)
\(\ge\frac{2}{1+\sqrt{a^3b^3}}+\frac{2}{1+\sqrt{abc^4}}\ge\frac{4}{1+\sqrt[4]{a^4b^4c^4}}=\frac{4}{1+abc}\)
PS: Đề sai nên t sửa luôn đề rồi nhé
\(\Rightarrow\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}\ge\frac{3}{1+abc}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2a-1}+\frac{1}{1}\ge\frac{4}{2a}=\frac{2}{a}\) ; \(\frac{1}{2b-1}+\frac{1}{1}\ge\frac{2}{b}\) ; \(\frac{1}{2c-1}+\frac{1}{1}\ge\frac{2}{c}\)
\(\Rightarrow VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\Rightarrow VT\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\sqrt{a^3+1}}=\frac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\ge\frac{2}{a+1+a^2-a+1}=\frac{2}{a^2+2}\)
Thiết lập tương tự: \(\frac{1}{\sqrt{b^3+1}}\ge\frac{2}{b^2+2}\) ; \(\frac{1}{\sqrt{c^3+1}}\ge\frac{2}{c^2+2}\)
\(\Rightarrow VT\ge\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}=\frac{1}{\frac{a^2}{2}+1}+\frac{1}{\frac{b^2}{2}+1}+\frac{1}{\frac{c^2}{2}+1}\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=\frac{1}{8}\)
\(\Rightarrow VT\ge\frac{x^2}{x^2+\frac{1}{2}}+\frac{y^2}{y^2+\frac{1}{2}}+\frac{z^2}{z^2+\frac{1}{2}}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+\frac{3}{2}}\)
\(\Rightarrow VT\ge\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x^2+y^2+z^2+\frac{3}{2}}\ge\frac{x^2+y^2+z^2+6.\sqrt[3]{\left(xyz\right)^2}}{x^2+y^2+z^2+\frac{3}{2}}=\frac{x^2+y^2+z^2+\frac{3}{2}}{x^2+y^2+z^2+\frac{3}{2}}=1\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) hay \(a=b=c=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trước hết, ta chứng minh bổ đề sau: Nếu \(a,b\ge1\)thì \(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(\frac{1}{1+a}-\frac{1}{1+\sqrt{ab}}\right)+\left(\frac{1}{1+b}-\frac{1}{1+\sqrt{ab}}\right)\ge0\)\(\Leftrightarrow\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)}+\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\)\(\Leftrightarrow\frac{\sqrt{b}\left(1+a\right)\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{a}\left(1+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\)\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\)*đúng do \(\sqrt{ab}\ge1\)(vì a,b\(\ge1\))*
Áp dụng bổ đề trên, ta được: \(\left(\frac{1}{1+a^4}+\frac{1}{1+b^4}\right)+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
Tương tự: \(\left(\frac{1}{1+b^4}+\frac{1}{1+c^4}\right)+\frac{2}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\left(\frac{1}{1+c^4}+\frac{1}{1+a^4}\right)+\frac{2}{1+a^4}\ge\frac{4}{1+ca^3}\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT C-S dạng ENgel ta có:
$$\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a} \ge \frac{3}{3+abc} $$
$$\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a} \ge \frac{9}{4(a+b+c)} $$
Ta chứng minh $$ \frac{9}{4(a+b+c)} \ge \frac{3}{3+abc} $ hay $9+3abc \ge 4(a+b+c) $$
Đặt $ a= 1-x, b=1-y, c=1-z $ rồi xài AM-GM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
Lời giải:
Ta nhớ đến 1 BĐT quen thuộc:
Với $x,y$ là số thực thỏa mãn $x,y\geq 1$ thì $\frac{1}{x^2+1}+\frac{1}{y^2+1}\geq \frac{2}{xy+1}$
(Việc chứng minh BĐT trên có thể dựa vào biến đổi tương đương)
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Áp dụng vào bài toán:
\(\frac{1}{a^3+1}+\frac{1}{b^3+1}\geq \frac{2}{\sqrt{a^3b^3}+1}(1)\)
\(\frac{1}{c^3+1}+\frac{1}{abc+1}\geq \frac{2}{\sqrt[2]{c^4ab}+1}(2)\)
\(\frac{1}{\sqrt{a^3b^3}+1}+\frac{1}{\sqrt{c^4ab}+1}=\frac{1}{[(ab)^{\frac{3}{4}}]^2+1}+\frac{1}{[c(ab)^{\frac{1}{4}})]^2+1}\geq \frac{2}{(ab)^{\frac{3}{4}}.c(ab)^{\frac{1}{4}}+1}=\frac{2}{abc+1}(3)\)
Từ \((1);(2);(3)\Rightarrow \frac{1}{a^3+1}+\frac{1}{b^3+1}+\frac{1}{c^3+1}+\frac{1}{abc+1}\geq \frac{4}{abc+1}\)
\(\Rightarrow \frac{1}{a^3+1}+\frac{1}{b^3+1}+\frac{1}{c^3+1}\geq \frac{3}{abc+1}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
@No choice teen