Chứng minh:
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\le\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\) (a,b,c>0)
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Áp dụng bất đẳng thức AM - GM ta có:
\(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{2}{2\sqrt{ab}}+\dfrac{2}{2\sqrt{bc}}+\dfrac{2}{2\sqrt{ac}}\)
\(=\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\le\dfrac{1}{\sqrt{a^2}}+\dfrac{1}{\sqrt{b^2}}+\dfrac{1}{\sqrt{c^2}}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu " = " xảy ra khi \(a=b=c\)
Vậy...
Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\). Tương tự cho 2 BĐT còn lại có:
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
Đẳng thức xảy ra khi \(a=b=c\)
Ta có đánh giá sau với a không âm:
\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)
Thật vậy, BĐT tương đương:
\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)
\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)
Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)
Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)
Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)
\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)
\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)
Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)
Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)
a.
Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
b.
Ta có:
\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)
Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
Cộng vế với vế:
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng BĐT Cauchy ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Làm tắt vài chỗ thông cảm
Câu b,
Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)
Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)
\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)
Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
a/(b+c) + b/(a+c) + c/(a+b) = a^2/(ab+ac) + b^2/(ba+bc) + c^2/(ac+bc) >=
(a+b+c)^2/(2.(ab+bc+ac) (buhihacopxki dạng phân thức)
>= (3.(ab+bc+ac)/(2(ab+bc+ac) =3/2
a^2/(b^2+c^2) + b^2/(a^2+c^2) + c^2/(a^2+b^2) >= (a+b+c)^2/(2.(a^2+b^2+c^2) (buhihacopxki dạng phân thức)
>= 3(a^2+b^2+c^2) / 2(a^2+b^2+c^2) >=3/2
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}-\dfrac{3}{2}\ge0\)
\(\Leftrightarrow\left(\dfrac{a}{b+c}-\dfrac{1}{2}\right)+\left(\dfrac{b}{c+a}-\dfrac{1}{2}\right)+\left(\dfrac{c}{a+b}-\dfrac{1}{2}\right)\ge0\)
\(\Leftrightarrow\left(\dfrac{2a-b-c}{2\left(b+c\right)}\right)+\left(\dfrac{2b-a-c}{2\left(a+c\right)}\right)+\left(\dfrac{2c-a-b}{2\left(a+b\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b+a-c}{2\left(b+c\right)}+\dfrac{b-a+b-c}{2\left(a+c\right)}+\dfrac{c-a+c-b}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(b+c\right)}+\dfrac{a-c}{2\left(b+c\right)}+\dfrac{b-a}{2\left(a+c\right)}+\dfrac{b-c}{2\left(a+c\right)}+\dfrac{c-a}{2\left(a+b\right)}+\dfrac{c-b}{2\left(a+b\right)}\ge0\)\(\Leftrightarrow\left(a-b\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+c\right)}\right]+\left(a-c\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]+\left(b-c\right)\left[\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\)
ta có: a,b,c là 3 số dương bất kì nên ta giả sử \(a\ge b\ge c\)
\(\Rightarrow a+c\ge b+c\)
\(\Leftrightarrow2\left(a+c\right)\ge2\left(b+c\right)\)
\(\Leftrightarrow\dfrac{1}{2\left(a+c\right)}\le\dfrac{1}{2\left(b+c\right)}\)
\(\Leftrightarrow\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(b+c\right)}\ge0\)
Mà \(a\ge b\Rightarrow a-b\ge0\)
\(\Rightarrow\left(a-b\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+c\right)}\right]\ge0\left(1\right)\)
Chứng minh tương tự, ta có:
\(\left(a-c\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\left(2\right)\)
\(\left(b-c\right)\left[\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\left(3\right)\)
Cộng từng vế (1);(2);(3) \(\Rightarrow\) luôn đúng
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)
\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)
\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)
\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)
Xảy ra khi \(a=b=c\)
\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)
\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)
Giả sử \(a\ge b\ge c\)thì
\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh
\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)
\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)
\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)
\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))
Vậy BĐT được chứng minh.
Dấu = xảy ra khi a=b=c
Áp dụng BĐT BSC:
\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(=\dfrac{b\left(a+b\right)-b^2}{a+b}+\dfrac{c\left(b+c\right)-c^2}{b+c}+\dfrac{a\left(c+a\right)-a^2}{c+a}\)
\(=a+b+c-\left(\dfrac{a^2}{c+a}+\dfrac{b^2}{a+b}+\dfrac{c^2}{c+a}\right)\)
\(\ge a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Đẳng thức xảy ra khi \(a=b=c\)
4ab ≤ (a + b)2 ⇒ \(\dfrac{4ab}{a+b}\le a+b\)
Tương tự \(\dfrac{4ac}{a+c}\le a+c\) ; \(\dfrac{4bc}{b+c}\le b+c\)
⇒ Cộng lại vế với vế :
4VT ≤ 2 (a+b+c) ⇒ VT ≤ \(\dfrac{a+b+c}{2}\)
\(BĐT\Leftrightarrow\left[\left(a+b\right)+\left(a+c\right)+\left(b+c\right)\right]\left(\dfrac{a^2+b^2}{a+b}+\dfrac{a^2+c^2}{a+c}+\dfrac{b^2+c^2}{b+c}\right)\le6\left(a^2+b^2+c^2\right)\)
Giả sử \(a\ge b\ge c\) thì \(a+b\ge a+c\ge b+c\) (**)
Và \(\dfrac{a^2+b^2}{a+b}\ge\dfrac{a^2+c^2}{a+c}\ge\dfrac{b^2+c^2}{b+c}\)(*)
Ta sẽ chứng minh (*) : \(\dfrac{a^2+b^2}{a+b}\ge\dfrac{a^2+c^2}{a+c}\Leftrightarrow ab\left(b-a\right)+ac\left(a-c\right)+bc\left(b-c\right)\ge0\)
\(\Leftrightarrow\left(b-c\right)\left[bc+a\left(b+c-a\right)\right]\ge0\)( đúng khi a,b,c là 3 cạnh 1 tam giác )
Tương tự :\(\dfrac{a^2+c^2}{a+c}\ge\dfrac{b^2+c^2}{b+c}\)
Từ (**) và (*) , Áp dụng BĐT chebyshev:( 2 dãy cùng chiều)
\(\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{a^2+b^2}{a+b}+\dfrac{a^2+c^2}{a+c}+\dfrac{b^2+c^2}{b+c}\right)\le3\left(a^2+b^2+b^2+c^2+c^2+a^2\right)=6\left(a^2+b^2+c^2\right)\)(đpcm)
Dấu = xảy ra khi a=b=c
bài này t lại quy đồng hết ra :v lười nghĩ quá :v Xem câu hỏi
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
Ta có \(\dfrac{a^2}{b^2}+1\ge2.\dfrac{a}{b}\)
Lập 2 BĐT tương tự rồi cộng theo vế, ta được:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (*)
Mà ta lại có \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=3\)
\(\Leftrightarrow-3\ge-\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (**)
Cộng theo vế (*) và (**), ta được đpcm.
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)