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Ta sẽ dùng phép biến đổi tương đương nhé :
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(1)
\(\Leftrightarrow\dfrac{b+a}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(b+a\right)}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\)
Vì a,b là các số dương =) ab(a+b) > 0
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2+2ab-4ab+b^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0 \)(luôn đúng) (2)
BĐT (2) luôn đúng mà các phép biến đổi trên là tương đương suy ra BĐT (1) đúng
Dấu "=" xảy ra khi và chỉ khi a = b
ta có \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}< =>\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(a+b\right)\ge4\)
<=>\(1+\dfrac{a}{b}+\dfrac{b}{a}+1\ge4\)
Thật vậy:
áp dụng bdt Cô si
=>\(1+\dfrac{a}{b}+\dfrac{b}{a}+1=2+\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2=4\)
vậy bất đăng thức xảy ra
dấu "=" xảy ra \(\Leftrightarrow\)a=b
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{b+c+a+c+a+b}\)
\(=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=VP\)
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
1) xét hiệu
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\ge0\)
<=> \(\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}-\dfrac{4ab}{ab\left(a+b\right)}\ge0\)
=> b(a+b)+a(a+b)-4ab ≥ 0
<=> ab+b2+a2+ab-4ab ≥ 0
<=> a2 -2ab+b2 ≥ 0
<=> (a-b)2 ≥ 0 (luôn đúng )
=> đpcm
2)Ta có:\(\left(a-b\right)^2\ge0\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
TT\(\Rightarrow\left(b+c\right)^2\ge4bc;\left(c+a\right)^2\ge4ca\)
\(\Rightarrow\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\ge64a^2b^2c^2\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
AM-GM: \(\left\{{}\begin{matrix}\dfrac{a^2}{b}+b\ge2a\\\dfrac{b^2}{c}+c\ge2b\\\dfrac{c^2}{a}+a\ge2c\end{matrix}\right.\) Cộng theo vế suy ra đpcm. Dấu "=" khi \(a=b=c\)
thank nha