\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

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

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)

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

\(\Rightarrow\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}\ge\frac{3}{1+abc}\)

khó quá bạn ơi mình cần thêm thời gian để làm 

nhanh lên nhé

Ta có \(a+b+b+b\ge4\sqrt[4]{abbb}\)(theo BĐT Cosi)

\(\Leftrightarrow a+3b\ge\sqrt[4]{ab^3}\)

\(\Leftrightarrow\frac{a+3b}{4}\ge4\sqrt[4]{ab^3}\)

Mà \(a,b,c\ge1\Rightarrow a+3b\ge4\Rightarrow\frac{a+3b}{4}\ge1\)

\(\Leftrightarrow1+\sqrt[4]{ab^3}\ge1+a\)

\(\Rightarrow\frac{1}{1+\sqrt[4]{ab^3}}\le\frac{1}{1+a}\left(1\right)\)

Tương tự \(\hept{\begin{cases}\frac{1}{1+\sqrt[4]{bc^3}}=\frac{1}{1+b}\left(2\right)\\\frac{1}{1+\sqrt[4]{ca^3}}=\frac{1}{1+c}\left(3\right)\end{cases}}\)

(1) (2) (3) => \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{1}{1+\sqrt[4]{ab^3+1}}+\frac{1}{1+\sqrt[4]{bc^3}}+\frac{1}{1+\sqrt[4]{ca^3}}\)(đpcm)

Bài 1:

a)Áp dụng Bđt Bunhiacopski ta có:

\(3a^2+4b^2\ge\frac{\left(3a+4b\right)^2}{7}=7\)

b)Áp dụng Bđt Bunhiacopski ta có:

\(\left(3a^2+5b^2\right)\left[\left(\frac{2}{\sqrt{3}}\right)^2+\left(-\frac{3}{\sqrt{5}}\right)^2\right]\ge\left(2a-3b\right)^2=49\)

\(\Rightarrow3a^2+5b^2\ge\frac{735}{47}\)

c)Áp dụng Bđt Bunhiacopski ta có:

\(\left(7a^2+11b^2\right)\left[\left(\frac{3}{\sqrt{7}}\right)^2+\left(\frac{5}{\sqrt{11}}\right)^2\right]\ge\left(\frac{3}{\sqrt{7}}\cdot\sqrt{7}a-\frac{5}{\sqrt{11}}\cdot\sqrt{11}b\right)^2=64\)

\(\Rightarrow\frac{274}{77}\left(7a^2+11b^2\right)\ge64\)

\(\Rightarrow7a^2+11b^2\ge\frac{2464}{137}\)

d)Áp dụng Bđt Bunhiacopski ta có:

\(\left(1^2+2^2\right)\left(a^2+b^2\right)\ge\left(a+2b\right)^2=4\)

\(\Rightarrow a^2+b^2\ge\frac{4}{5}\)

lần sau đăng ít thôi nhé

a)Áp dụng Bđt Bunhiacopski ta có:

\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)

\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)

phần khác tương tư

\(\frac{1}{2a-1}+\frac{1}{1}\ge\frac{4}{2a}=\frac{2}{a}\) ; \(\frac{1}{2b-1}+\frac{1}{1}\ge\frac{2}{b}\) ; \(\frac{1}{2c-1}+\frac{1}{1}\ge\frac{2}{c}\)

\(\Rightarrow VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)\)

\(\Rightarrow VT\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

1.

C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)

\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)

\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)

\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)

\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)

Theo bài ra: \(a^3-b^3\ge3a-3b-4\)

\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)

Thật vậy:

\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)

\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)

\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)

\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)

\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))

2.

\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)

\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)

\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)

\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)

\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)

\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)