cho ba số thực dương a,b,c.Chứng minh rằng:
\(\dfrac{a^2+b^2}{a+b}+\dfrac{c^2+b^2}{c+b}+\dfrac{c^2+a^2}{c+a}\le\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
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\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}-2+\dfrac{c\left(2b-c\right)}{b\left(c+a\right)}-2+\dfrac{a\left(2c-a\right)}{c\left(a+b\right)}-2\le\dfrac{3}{2}-6\)
\(\Leftrightarrow\dfrac{b^2+2ac}{a\left(b+c\right)}+\dfrac{c^2+2ab}{b\left(c+a\right)}+\dfrac{a^2+2bc}{c\left(a+b\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{b^2}{ab+ac}+\dfrac{c^2}{bc+ab}+\dfrac{a^2}{ac+bc}+\dfrac{2c^2}{bc+c^2}+\dfrac{2a^2}{ac+a^2}+\dfrac{2b^2}{ab+b^2}\ge\dfrac{9}{2}\)
Ta có:
\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}+\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)
\(\Leftrightarrow VT\ge\left(a+b+c\right)^2\left(\dfrac{1}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}\right)\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a^2+b^2+c^2+ab+bc+ca\right)}\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{9}{2}\)
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
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)
\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)
Nhìn qua đã biết là đề sai rồi bạn
Cho \(a,b,c\) các giá trị lớn ví dụ \(a=b=c=2\) là thấy sai ngay
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
Cho 3 số dương a, b, c. Chứng minh:\(\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le3\left(\dfrac{a^2+... - Hoc24
tham khảo ^^