Cho ba số a, b, c thỏa mãn: b ≠ c và a + b ≠ c và c2 = 2(ac + bc - ab)
Chứng minh rằng: \(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{a-c}{b-c}\)
Cho \(a,b,c\) thỏa mãn \(\left|a\right|,\left|b\right|,\left|c\right|< 1\) và \(ab+bc+ca=2\). Chứng minh :
\(P=\dfrac{a^2}{1-b^2}+\dfrac{b^2}{1-c^2}+\dfrac{c^2}{1-a^2}\ge6\).
\(a^2+b^2+c^2\ge ab+bc+ca=2\)
Áp dụng BĐT C-S:
\(P\ge\dfrac{\left(a+b+c\right)^2}{3-\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2+4}{3-\left(a^2+b^2+c^2\right)}\)
Đặt \(a^2+b^2+c^2=x\)
Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)
\(\Leftrightarrow x\ge2\) (đúng)
Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)
Cho a,b,c thỏa mãn ab+bc+ca =1. Chứng minh rằng
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Ta có VP:
\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Thay \(1=ab+bc+ca\)
\(=\dfrac{2}{\sqrt{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}}\)
\(=\dfrac{2}{\sqrt{\left[b\left(a+c\right)+a\left(a+c\right)\right]\left[a\left(b+c\right)+b\left(b+c\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]}}\)
\(=\dfrac{2}{\sqrt{\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(a+c\right)}}\)
\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)
\(=\dfrac{2}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)
_____________
Ta có VT:
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)
Thay \(1=ab+ac+bc\)
\(=\dfrac{a}{ab+ac+bc+a^2}+\dfrac{b}{ab+ac+bc+b^2}+\dfrac{c}{ab+ac+bc+c^2}\)
\(=\dfrac{a}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b}{b\left(b+c\right)+a\left(b+c\right)}+\dfrac{c}{c\left(b+c\right)+a\left(b+c\right)}\)
\(=\dfrac{a}{\left(a+c\right)\left(a+b\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{a\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}+\dfrac{b\left(a+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\dfrac{c\left(a+b\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{ab+ac+ab+bc+ac+bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2ab+2ac+2bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2\cdot\left(ab+ac+bc\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\left(ab+ac+bc=1\right)\)
Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\left(dpcm\right)\)
Cho 3 số a, b, c thỏa mãn a # -b, b # -c, c # -a.
Chứng minh rằng : \(\dfrac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^2-ab}{\left(c+a\right)\left(c+b\right)}=0\)
Cho 3 số thực dương a,b,c thoả mãn: a2+b2+c2=1 .Chứng minh:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}+\dfrac{b^5+c^5}{bc\left(b+c\right)}+\dfrac{c^5+a^5}{ca\left(c+a\right)}\ge3\left(ab+bc+ca\right)-2\)
Ta chứng minh BĐT sau cho các số dương:
\(x^5+y^5\ge xy\left(x^3+y^3\right)\)
\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
Áp dụng:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)
Tương tự và cộng lại:
\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)
\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)
\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)
Cho các số thực dương : \(a;b;c\) thỏa mãn điều kiện : \(ab+bc+ac+abc=4\)
Chứng minh rằng : \(\dfrac{1}{\sqrt{2.\left(a^2+b^2\right)}+4}+\dfrac{1}{\sqrt{2.\left(b^2+c^2\right)}+4}+\dfrac{1}{\sqrt{2.\left(c^2+a^2\right)}+4}\le\dfrac{1}{2}\)
P/s: Em xin phép nhờ sự giúp đỡ của quý thầy cô giáo và các bạn yêu toán.
Em cám ơn nhiều lắm ạ!
Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)
Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\)
Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)
" = " \(\Leftrightarrow a=b=c=1\)
1.Cho \(a,b,c,d\) là các số nguyên thỏa mãn \(a^3+b^3=2\left(c^3-d^3\right)\) . Chứng minh rằng a+b+c+d chia hết cho 3
2.Cho ba số dương a,b,c thỏa mãn abc=1. Chứng minh rằng \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Cho a, b, c là ba số dương thỏa mãn \(abc\)=1. Chứng minh rằng:
\(\dfrac{1}{a^3\left(b+c\right)}\)+\(\dfrac{1}{b^3\left(a+c\right)}\)+\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
Cho ba số hữu tỉ a, b, c thỏa mãn: \(a.b.c=1\) và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Chứng minh rằng biểu thức \(A=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\) có giá trị bằng bình phương của một số hữu tỉ.
Thôi câu đó mình làm được rồi, các bạn giúp mình câu này nha
Cho \(a>b\ge0\). CMR: \(\dfrac{a^4+b^4}{a^4-b^4}-\dfrac{ab}{a^2-b^2}+\dfrac{a+b}{2\left(a-b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\\ \to ab+bc+ca=abc=1\)
Ta có \(A=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)
\(\to A=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(\to A=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Vì $a,b,c\in \mathbb{Q}\to A\in \mathbb{Q}$
cho ba số a,b,c là các số dương thoả mãn abc=1.chứng minh rằng:\(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+c+1\right)^2}+\dfrac{c}{\left(ac +c+1\right)^2}\ge\dfrac{1}{a+b+c}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\((ab+a+1)^2 \le (a+b+c) \left( a+ a^2b+ \frac 1c \right) = (a+b+c)(a+a^2b+ab)\)
\(\Rightarrow \dfrac{a}{(ab+a+1)^2} \ge \dfrac{a}{(a+b+c)(a+a^2b+ab)}= \dfrac{1}{(a+b+c)(1+ab+b)}\)
Thiết lập các BĐT tương tự rồi cộng theo vế ta có:
\(\sum \dfrac{a}{(ab+a+1)^2} \ge \dfrac{1}{a+b+c} \sum \dfrac{1}{ab+b+1}= \dfrac{1}{a+b+c}\)
c2: Áp dụng BĐT bunyakovsky:
\(\left(a+b+c\right)\left[\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right]\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\right)^2\)
Xét \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
\(=\dfrac{ab+a+1}{ab+a+1}=1\)
do đó \(\left(a+b+c\right).VT\ge1\Leftrightarrow VT\ge\dfrac{1}{a+b+c}\)
dấu = xảy ra khi a=b=c=1