Cho 3 số thực phân biệt a,b,c.Chứng minh\(a^4\left(b-c\right)+b^4\left(c-a\right)\ne c^4\left(b-a\right)\)
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29 tháng 3 2021
\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}-2+\dfrac{c\left(2b-c\right)}{b\left(c+a\right)}-2+\dfrac{a\left(2c-a\right)}{c\left(a+b\right)}-2\le\dfrac{3}{2}-6\)
\(\Leftrightarrow\dfrac{b^2+2ac}{a\left(b+c\right)}+\dfrac{c^2+2ab}{b\left(c+a\right)}+\dfrac{a^2+2bc}{c\left(a+b\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{b^2}{ab+ac}+\dfrac{c^2}{bc+ab}+\dfrac{a^2}{ac+bc}+\dfrac{2c^2}{bc+c^2}+\dfrac{2a^2}{ac+a^2}+\dfrac{2b^2}{ab+b^2}\ge\dfrac{9}{2}\)
Ta có:
\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}+\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)
\(\Leftrightarrow VT\ge\left(a+b+c\right)^2\left(\dfrac{1}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}\right)\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a^2+b^2+c^2+ab+bc+ca\right)}\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{9}{2}\)
17 tháng 3 2022
a) phương trình \(x^3-3x^2+1\) có 3 nghiệm thực phân biệt là a,b,c(đề bài). Áp dụng Định lí Vi-ét cho đa thức bậc 3 ta có:\(\left\{{}\begin{matrix}a+b+c=3\\ab+bc+ac=0\\a.b.c=-1\end{matrix}\right.\)
ta có
a+b+c=3
<=>\(\left(a+b+c\right)^2=9\)
<=>\(a^2+b^2+c^2+2ab+2bc+2ac=9\)
<=>\(a^2+b^2+c^2=9\)
<=>\(\left(a^2+b^2+c^2\right)^2=81\)
<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=81\)(1)
ta có ab+bc+ac=0
<=>\(\left(ab+bc+ac\right)^2=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2-2.1.3=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2=6\)(2)
Thay (2) vào (1) ta có \(a^4+b^4+c^4+2.6=81\)
<=>\(a^4+b^4+c^4=69\)
17 tháng 3 2022
b) \(\dfrac{a+1}{\left(b+c\right)\left(1-a\right)+1}=\dfrac{a+1}{\left(3-a\right)\left(1-a\right)+1}=\dfrac{a+1}{3+a^2-4a+1}=\dfrac{a+1}{a^2-4a+4}=\dfrac{a+1}{\left(a-2\right)^2}\)
cmtt =>\(B=\dfrac{a+1}{\left(a-2\right)^2}+\dfrac{b+1}{\left(b-2\right)^2}+\dfrac{c+1}{\left(c-2\right)^2}\)=\(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}+3\left[\dfrac{1}{\left(a-2\right)^2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right]\)=\(\dfrac{3\left[\left(a-2\right)\left(b-2\right)\right]^2+3\left[\left(b-2\right)\left(c-a\right)\right]^2+3\left[\left(c-2\right)\left(a-2\right)\right]^2}{\left[\left(a-2\right)\left(b-2\right)\left(c-2\right)\right]^2}\)
đặt t=(a-2)(b-2);u=(b-2)(c-2);v=(c-2)(a-2) =>t+u+v=0
B thành \(\dfrac{3\left(t^2+u^2+v^2\right)}{t.u.v}\) bạn biến đổi để xuất hiện t+u+v
=>B=\(\dfrac{3\left(t+u+v\right)^2-6\left(t.u+u.v+t.v\right)}{t.u.v}=\dfrac{-6.\left(a-2\right)\left(b-2\right)\left(c-2\right)\left(a-2+b-2+c-2\right)}{t.u.v}=\dfrac{18}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)
(a-2)(b-2)(c-2)= abc-2(ab+bc+ac)+4(a+b+c)-8=12-9=3
Vậy B=3
2 tháng 2 2018
Ta dự đoán :\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{a^2}{a^2+b^2+c^2}\)
Thật vậy ta sẽ chứng minh nó:
\(\Leftrightarrow\left(a^2+b^2+c^2\right)\ge a\left(a^3+\left(b+c\right)^3\right).\)
\(\Leftrightarrow2a^2\left(b^2+c^2\right)+\left(b^2+c^2\right)^2\ge a\left(b+c\right)^3\left(#\right)\)
Ta có:\(2a^2\left(b^2+c^2\right)+\left(b^2+c^2\right)^2\ge a^2\left(b+c\right)^2+\frac{1}{4}\left(b+c\right)^4\ge a\left(b+c\right)^3\)
Từ đó , ta có bất đẳng thức \(\left(#\right).\)
Tương tự:
\(\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}\ge\frac{b^2}{a^2+b^2+c^2}\)
\(\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge\frac{c^2}{a^2+b^2+c^2}.\)
Cộng bất đẳng thức trên lại ta có điểu phải chứng minh.
Dấu bằng xảy ra khi \(a=b=c\)
23 tháng 9 2021
Ta có: \(a+b+c+\sqrt{abc}=4\)
\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)
\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
23 tháng 9 2021
Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)
\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)
Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)
T
26 tháng 4 2020
\(\Leftrightarrow\left(\Sigma a\right)^4\left(\Sigma a^4b^4\right)\left[\Sigma c^2\left(a^2+b^2\right)^2\right]\ge54^2\left(abc\right)^6\)
Giả sử \(c=\text{min}\left\{a,b,c\right\}\)và đặt \(a=c+u,b=c+v\) thì nhận được một BĐT hiển nhiên :P
26 tháng 4 2020
Theo BĐT AM-GM ta có:
\(c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)\ge3\sqrt[3]{\left(abc\right)^2\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\right]^2}\)
\(\ge3\sqrt[3]{\left(abc\right)^264\left(abc\right)^4}=12\left(abc\right)^2\)
=> \(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(a^2+c^2\right)^2}\ge2\sqrt{3}abc\)
Cũng theo BĐT AM-GM \(\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4\ge3\sqrt[3]{\left(ab\right)^4\left(bc\right)^4\left(ca\right)^4}=3\left(abc\right)^2\sqrt[3]{\left(abc\right)^2}\)
=> \(\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}\ge\sqrt{3}\cdot abc\sqrt[3]{abc}\)và \(\left(a+b+c\right)^2\ge9\sqrt[3]{\left(abc\right)^2}\)
=> \(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)^2}\cdot\left(a+b+c\right)^2\cdot\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}\)
\(\ge2\sqrt{3}\left(abc\right)\cdot\sqrt{3}\left(abc\right)\sqrt[3]{abc}\cdot9\sqrt[3]{\left(abc\right)^2}\ge54\left(abc\right)^3\)
Dấu "=" xảy ra <=> a=b=c
H
25 tháng 5 2023
Theo giả thiết kết hợp sử dụng BĐT AM - GM có:
\(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-\left[c\left(a+b\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)
\(\le\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-2\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\left[\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\right]^2\)
Suy ra \(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\ge2\Leftrightarrow\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}\ge3\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge7\)
Khi đó, sử dụng BĐT Cauchy - Schwarz ta có:
\(\left(a^4+b^4+c^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\ge\left[\sqrt{\left(a^4+b^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}\right)}+1\right]^2\)
\(=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+1\right)^2=\left[\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-1\right]^2\ge\left(7^2-1\right)^2=2304\)
Đẳng thức xảy ra khi và chỉ khi \(ab=c^2\) và \(\dfrac{a}{b}+\dfrac{b}{a}=7\)
25 tháng 5 2023
(a+b-c)(1/a+1/b-c)=(a+b)(1/a+1/b)+1-[c(a+b)+c(1/a+1/b)]<=(a+b)(1/a+1/b)+1-2căn (a+b)(1/a+1/b)
=[(căn (a+b)(1/a+1/b))-1]^2
=>\(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1>=2\)
=>\(\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}>=3\)
=>a/b+b/a>=7
(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=[căn ((a^4+b^4)(1/a^4+1/b^4))+1]^2
=(a^2/b^2+b^2/a^2+1)^2=[(a/b+b/a)^2-1]^2>=(7^2-1)^2=2304
=>ĐPCM
Giả sử: \(a^4\left(b-c\right)+b^4\left(c-a\right)=c^4\left(b-a\right)\)
\(\Leftrightarrow a^4\left(b-a+a-c\right)+b^4\left(c-a\right)-c^4\left(b-a\right)=0\)
\(\Leftrightarrow a^4\left(b-a\right)+a^4\left(a-c\right)+b^4\left(c-a\right)-c^4\left(b-a\right)=0\)
\(\Leftrightarrow\left(b-a\right)\left(a^4-c^4\right)+\left(a-c\right)\left(a^4-b^4\right)=0\)
\(\Leftrightarrow\left(b-a\right)\left(a-c\right)\left(a+c\right)\left(a^2+c^2\right)+\left(a-c\right)\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)=0\)
\(\Leftrightarrow\left(b-a\right)\left(c-a\right)\left\{\left(a+c\right)\left(a^2+c^2\right)-\left(a+b\right)\left(a^2+b^2\right)\right\}=0\)
\(\Leftrightarrow\left(a+c\right)\left(a^2+c^2\right)-\left(a+b\right)\left(a^2+b^2\right)=0\)( do a, b, c phân biệt).
\(\Leftrightarrow ac^2+a^2c+c^3-ab^2-a^2b-b^3=0\)
\(\Leftrightarrow a^2\left(c-b\right)+a\left(c^2-b^2\right)+\left(c^3-b^3\right)=0\)
\(\Leftrightarrow\left(c-b\right)\left(a^2+a\left(b+c\right)+b^2+bc+c^2\right)=0\)
\(\Leftrightarrow\left(c-b\right)\left(a^2+2.a\frac{b+c}{2}+\frac{b^2+2bc+c^2}{4}+\frac{3b^2+2bc+3c^2}{4}\right)=0\)
\(\Leftrightarrow\left(c-b\right)\left(\left(a+\frac{b+c}{2}\right)^2+\frac{2b^2+3bc+2c^2}{4}\right)=0\)(*).
Do \(\left(a+\frac{b+c}{2}\right)^2\ge0,\frac{2b^2+3bc+2c^2}{4}>0\).
Nên (*) không thể xảy ra. Vậy điều giả sử sai, ta có đpcm.
Đặt A = a4(b - c) + b4(c - a) + c4(a - b) = a4(b - a + a - c) + b4(c - a) + c4(a - b) = a4(b - a) + a4(a - c) + b4(c - a) + c4(a - b)
= (a - b)(c4 - a4) + (a - c)(a4 - b4) = (a - b)(c - a)(c + a)(c2 + a2) + (a - c)(a - b)(a + b)(a2 + b2)
= (a - b)(a - c)[(a + b)(a2 + b2) - (c + a)(c2 + a2)] = (a - b)(a - c)(a3 + ab2 + a2b + b3 - c3 - a2c - ac2 - a3)
= (a - b)(a - c)[a2(b - c) + a(b2 - c2) + (b3 - c3)] = (a - b)(a - c)(b - c)[a2 + a(b + c) + b2 + bc + c2]
= (a - b)(a - c)(b - c)\(\frac{a^2+2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2}{2}\)
=\(\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left[\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\right]}{2}\)
Vì a,b,c là 3 số phân biệt nên A khác 0 <=> a4(b - c) + b4(c - a)\(\ne-c^4\left(a-b\right)=c^4\left(b-a\right)\)