BĐT cần chứng minh tương đương :

\(\sqrt{\dfrac{a^2+b^2}{2}}-\sqrt{ab}\ge\dfrac{a+b}{2}-\dfrac{2ab}{a+b}\)

\(\Leftrightarrow\dfrac{\dfrac{a^2+b^2}{2}-ab}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a+b\right)^2-4ab}{2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\right)\ge0\)

ta phải chứng minh;

\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\ge0\)

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

\(\Leftrightarrow a+b\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\)\(\Leftrightarrow2a+2b-\sqrt{2\left(a^2+b^2\right)}-2\sqrt{ab}\ge0\)

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

\(\Leftrightarrow\dfrac{\left(a+b\right)^2-2\left(a^2+b^2\right)}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a+b\right)^2-4ab}{a+b+2\sqrt{ab}}\ge0\)

\(\Leftrightarrow\dfrac{-\left(a-b\right)^2}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a-b\right)^2}{a+b+2\sqrt{ab}}\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right)\ge0\)

ta phải chứng minh

\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\ge0\)

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

\(\Leftrightarrow a+b+2\sqrt{ab}\le a+b+\sqrt{2\left(a^2+b^2\right)}\)

\(\Leftrightarrow2\sqrt{ab}\le\sqrt{2\left(a^2+b^2\right)}\Leftrightarrow\left(a-b\right)^2\ge0\)

\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)

\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)

\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ 

\(a^2+b^2+1\ge ab+a+b\)

<=> \(2a^2+2b^2+2\ge2ab+2a+2b\)

<=>\(\left(a^2+2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

<=>\(\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)

do (a-b)²>=0, (a-1)²>=0,(b-1)²>=0=>BĐT được c/m

dấu ''=''m xảy ra khi và chỉ khi a=b=1

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

mình nhầm một chút ở chỗ \(\left(a^2+2ab+b^2\right)thành\left(a^2-2ab+b^2\right)nhé\)

Cosi ngược dấu

Giả sử \(a\ge b\ge c\)

Ta có:\(\frac{a+b}{ab+c^2}+\frac{b+c}{bc+a^2}+\frac{c+a}{ca+b^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Leftrightarrow\frac{ac+bc-ab-c^2}{c\left(ab+c^2\right)}+\frac{ab+ac-bc-a^2}{\left(bc+a^2\right)a}+\frac{cb+ab-ca-b^2}{b\left(ca+b^2\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}+\frac{\left(c-b\right)\left(b-a\right)}{b\left(ca+b^2\right)}\le0\)

Ta có:\(\left(c-b\right)\left(b-a\right)\ge0;\left(b-a\right)\left(a-c\right)\le0;\left(a-c\right)\left(c-b\right)\le0\)

\(\Rightarrow\frac{\left(c-b\right)\left(c-a\right)}{b\left(ca+b^2\right)}\le\frac{\left(c-b\right)\left(c-a\right)}{c\left(ab+c^2\right)}\)

\(\Rightarrow LHS\le\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(c-b\right)\left(b-a\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\)

\(=\frac{-\left(c-b\right)^2}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)c}\le0\)

\(\Rightarrowđpcm\)

Trả lời :

Bn Thu Trang đừng bình luận linh tinh nhé.

- Hok tốt !

^_^

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

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

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow a^2+b^2+a^3b+ab^3+2ab+2\ge2a^2b^2+2a^2+2b^2+2\)

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

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

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

Vì bđt cuối luôn đúng với mọi \(a\ge1;b\ge1\) mà các biến đổi trên là tương đương nên bđt đầu luôn đúng

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

Trời ko ai giải dùm hả 

Thôi chắc mình tự trả lời cho mn tham khảo quá.

Áp dụng BĐT Cauchy dạng :\(\frac{x+y}{2}\ge\sqrt{x+y}\Leftrightarrow x+y\ge2\sqrt{xy}\) 

                        Dấu "=" xảy ra khi : x = y

Ta có : 

          \(ab+\frac{a}{b}\ge2.\sqrt{ab.\frac{a}{b}}=2\sqrt{a^2}=2a\)

Tương tự : \(\frac{a}{b}+\frac{b}{a}\ge2\)

                    \(ab+\frac{b}{a}\ge2b\)

Cộng vế với vế ta được : 

       \(2\left(ab+\frac{a}{b}+\frac{b}{a}\right)\ge2\left(a+b+1\right)\)

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

BĐT \(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge a+b+c+ab+bc+ca\)

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{1}{4}\left(y+z-x\right)^2+a^2+b^2+c^2-\left(a+b+c\right)\ge0\)

Có: \(VT\ge\frac{3}{4}\left(y-z\right)^2+\frac{1}{4}\left(y+z-x\right)^2+\left[\frac{\left(a+b+c\right)^2}{3}-\left(a+b+c\right)\right]\ge0\)(chú ý: \(\left(a+b+c\right)^2=\left(a+b+c\right)\left(a+b+c\right)\ge3\sqrt[3]{abc}\left(a+b+c\right)=3\left(a+b+c\right)\))

Ta có đpcm.

Có cách khác ^_^ mới nghĩ ra

BĐt <=> \(P\left(a,b,c\right)=a^2+b^2+c^2-\frac{1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge0\)

Không mất tính tổng quát , giả sử : \(a=min\left\{a,b,c\right\}\Rightarrow t=\sqrt{bc}\ge1\)

=> Chứng minh: \(P\left(a,b,c\right)\ge P\left(a,t,t\right)\)

Thật vậy , \(P\left(a,b,c\right)-P\left(a,t,t\right)=\left(\sqrt{b}-\sqrt{c}\right)^2\left[\left(\sqrt{b}+\sqrt{c}\right)^2-\frac{1}{2}\left(1+\frac{1}{bc}\right)\right]\)

                                                                  \(\ge\left(\sqrt{b}-\sqrt{c}\right)^2\left[4-\frac{1}{2}\left(1+1\right)\right]\ge0\)

mặt khác: \(P\left(a,t,t\right)=P\left(\frac{t}{t^2},t,t\right)=\frac{\left(t-1\right)^2\left(3t^4+4t^3+5t^2+4t+2\right)}{2t^4}\ge0\)

=> BĐT được chứng minh . Đt xảy ra<=> a=b=c=1

Nguyễn Văn Tuấn Anh cách đó phức tạp lắm, nên mình không làm;))

\(bdt\Leftrightarrow a^2+b^2+c^2-ab-ac-bc-\frac{\left(a+b\right)^2}{26}-\frac{\left(b-c\right)^2}{6}-\frac{\left(c-a\right)^2}{2009}\ge0\)

\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]-\frac{\left(a+b\right)^2}{26}-\frac{\left(b-c\right)^2}{6}-\frac{\left(c-a\right)^2}{2009}\ge0\)

Đặt \(a-b=x;b-c=y;c-a=z\) nên

\(bdt\Leftrightarrow\frac{1}{2}\left(x^2+y^2+z^2\right)-\frac{x^2}{26}-\frac{y^2}{6}-\frac{z^2}{2009}\ge0\)

\(\Leftrightarrow\left(\frac{x^2}{2}-\frac{x^2}{26}\right)+\left(\frac{y^2}{2}-\frac{y^2}{6}\right)+\left(\frac{z^2}{2}-\frac{z^2}{2009}\right)\ge0\)

\(\Leftrightarrow\frac{6x^2}{13}+\frac{y^2}{3}+\frac{2007z^2}{4018}\ge0\)(luôn đúng \(\forall x;y;z\))

Vậy BTĐ đã được chứng minh

ta cần chứng minh a^2+b^2+1-ab-a-b>hoặc bằng 0

2(a^2+b^2+1-ab-a-b)>hoặc bằng 0

2a^2+2b^2+2-2ab-2a-2b>hoặc bằng 0

a^2-2ab+b^2+a^2-2a+1+b^2-2b+1>hoặc bằng 0

(a-b)^2+(a-1)^2+(b-1)^2>hoặc bằng 0 (luôn đúng)

suy ra pt trên đúng