Sửa đề:  Cho a, b, c là các số thực dương thỏa mãn điều kiện abc=1. Chứng minh rằng

\(\frac{1}{ab+b+2}+\frac{1}{bc+c+2}+\frac{1}{ca+a+2}\le\frac{3}{4}\)

Áp dụng bđt Cauchy-Schwarz ta có:

\(\frac{1}{ab+b+2}=\frac{1}{ab+1+b+1}\le\frac{1}{4}\left(\frac{1}{ab+1}+\frac{1}{b+1}\right)\) \(=\frac{1}{4}\left(\frac{abc}{ab\left(1+c\right)}+\frac{1}{b+1}\right)=\frac{1}{4}\left(\frac{c}{1+c}+\frac{1}{b+1}\right)\)

Tương tự \(\frac{1}{bc+c+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{c+1}\right)\)

          \(\frac{1}{ca+a+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{a+1}\right)\)

Cộng từng vế các bđt trên ta được

\(VT\le\frac{1}{4}\left(\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\right)=\frac{3}{4}\)

Vậy bđt được chứng minh

Dấu "=" xảy ra khi a=b=c=1

vì \(a+b+c=1\)

\(< =>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)

\(=3+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}+\frac{a}{c}\)

\(=3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)

ta có pt:

\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\right)\)

\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{3}{4}+\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\)

áp dụng bđt cô- si( cauchy) gọi pt là P 

\(P\ge2\sqrt{\frac{ab}{a^2+b^2}\frac{a^2+b^2}{4ab}}+2\sqrt{\frac{bc}{b^2+c^2}\frac{b^2+c^2}{4bc}}+2\sqrt{\frac{ca}{c^2+a^2}\frac{c^2+a^2}{4ca}}+\frac{3}{4}\)

\(P\ge2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}\)

\(P\ge2.\frac{1}{2}+2.\frac{1}{2}+2.\frac{1}{2}+\frac{3}{4}\)

\(P\ge1+1+1+\frac{3}{4}=\frac{15}{4}\)

dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)

<=>ĐPCM

Svacxo chăng :33 Ai thử đi, e sợ biến nhiều lắm :))

\(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\Leftrightarrow\frac{1}{1+a}+\frac{1}{1+b}-\frac{2}{1+\sqrt{ab}}\ge0\)

\(\Leftrightarrow\left(\frac{1}{a+1}-\frac{1}{1+\sqrt{ab}}\right)+\left(\frac{1}{b+1}-\frac{1}{1+\sqrt{ab}}\right)\ge0\)

\(\Leftrightarrow\frac{\sqrt{ab}-a}{\left(a+1\right)\left(1+\sqrt{ab}\right)}+\frac{\sqrt{ab}-b}{\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)

\(\Leftrightarrow\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\left(a+1\right)\left(1+\sqrt{ab}\right)}+\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)

\(\Leftrightarrow\frac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(b+1\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\left(a+1\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)

\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a\sqrt{b}+\sqrt{b}-b\sqrt{a}-\sqrt{a}\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)

\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)(đúng với \(ab\ge1\))

Vậy \(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)

 Đẳng thức xảy ra khi a = b 

Bài này: nên đặt a=x^2; b=y^2

Nội suy  đỡ đau đầu hơn.

\(\sqrt[4]{b^3}\)

Vì a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{a}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)=2+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)

Do đó

\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{ab}\right)+\left(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{bc}\right)+\left(\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}\right)+\frac{3}{4}\)

\(\ge2\sqrt{\frac{ab}{a^2+b^2}\cdot\frac{a^2+b^2}{ab}}+2\sqrt{\frac{bc}{c^2+b^2}\cdot\frac{c^2+b^2}{bc}}+2\sqrt{\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}}+\frac{3}{4}\)

\(=2\cdot\frac{1}{2}+2\cdot\frac{1}{2}+\frac{2}{3}=\frac{15}{4}\)

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

ta sẽ giết ngươi kí tên dép đờ kiu lờ

Nhân 2 vế của \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) có: \(ab+bc+ca=abc\)

Ta có: 

\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^2}{a+bc}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\)

\(\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(a+c\right)}\cdot\frac{a+b}{8}\cdot\frac{a+c}{8}}=\frac{3a}{4}\)

Tương tự cho 2 BĐT còn lại ta có:

\(\frac{b^2}{b+ca}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3b}{4};\frac{c^2}{c+ab}+\frac{a+c}{8}+\frac{b+c}{8}\ge\frac{3c}{4}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\frac{4\left(a+b+c\right)}{8}\ge\frac{3\left(a+b+c\right)}{4}\)

\(\Leftrightarrow VT+\frac{4\left(a+b+c\right)}{8}\ge\frac{6\left(a+b+c\right)}{8}\)

\(\Leftrightarrow VT\ge\frac{a+b+c}{4}=VP\). Ta có ĐPCM

Bài 2 : đã cm bên kia

Bài 1: :| 

we had điều này:

\(2=\frac{2014}{x}+\frac{2014}{y}+\frac{2014}{z}\)

\(\Leftrightarrow\frac{x-2014}{x}+\frac{y-2014}{y}+\frac{z-204}{z}=1\)

Xòng! bunyakovsky

P/s : Bệnh lười kinh niên tái phát nên ít khi ol sorry :<

xét hiệu \(\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\)

quy đồng làm nốt nha