BĐT tương đương vs
(\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\))^2\(\ge\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+b^2+c^2+d^2+2ac+2bd\)
\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)( BĐT bunyakovsky ) luôn đúng
\(\Rightarrow\) đpcm
Chm: \(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge\left(a+b\right)\left(c+d\right)\) voi \(a,b,c,d>0\)
Chm bdt sau bang phuong phap hinh hoc:
\(\sqrt{a^2+b^2}.\sqrt{b^2+c^2}\ge b\left(a+c\right)\) (a,b,c>0)
\(a,b,c,d\in R\). CM :
\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{2}\left(\sqrt[4]{\left(a+b\right)^2\left|c+d\right|}-\sqrt[4]{\left|a+b\right|\left(c+d\right)^2}\right)\)
Với mọi a, b, c, x, y, z \(\in\) R, chứng minh : \(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b+c\right)^2+\left(x+y+z\right)^2}\)
câu 1 :
Cho 3 số x,y,z thỏa mãn 0<x,y,z≤1 và x+y+z=2
Tìm GTNN của \(A=\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\)
câu 2 :
Tìm giá trị lớn nhất của A
Với a,b,c , d là các số dương và \(a+b+c+d\le1\)
\(A=\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}+\sqrt{c}\right)^4+\left(\sqrt{a}+\sqrt{d}\right)^4+\left(\sqrt{b}+\sqrt{c}\right)^4+\left(\sqrt{b}+\sqrt{d}\right)^4+\left(\sqrt{c}+\sqrt{d}\right)^4\)
Cho a,b,c > 0 , \(a^2+b^2+c^2=3\). Chứng minh rằng : \(\sqrt{\frac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\frac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\frac{9}{\left(a+c\right)^2}+b^2}\)≥\(\frac{3\sqrt{13}}{2}\)
Tính:
\(a.\) \(A=\sqrt{12}-2\sqrt{48}+\dfrac{7}{5}\sqrt{75}\)
\(b.\) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(c.\) \(C=\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
\(d.\) \(D=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}-5}{\sqrt{5}}-\dfrac{11}{2\sqrt{5}+3}\)
1. Cho \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) , trong đó: \(x,y,z>0\)
Chm: \(x=y=z\)
2. Cho \(a_1,a_2,...,a_n>0\) và \(a_1a_2...a_n=1\) Chm: \(\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right)\ge2^n\)
3. Chm \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge2\sqrt{2\left(a+b\right)\sqrt{ab}}\) \(\left(a,b\ge0\right)\)
Cho a,b,c>0 thỏa mãn: a.b.c=8
Chứng minh: \(\frac{a^2}{\sqrt{\left(1+a^3\right).\left(1+b^3\right)}}+\frac{b^2}{\sqrt{\left(1+b^3\right).\left(1+c^3\right)}}+\frac{c^2}{\sqrt{\left(1+c^3\right).\left(1+a^3\right)}}\ge\frac{4}{3}\)