Áp dụng BĐT Bunhia:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)}\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3.\left(4\left(a+b+c\right)+3\right)}=\sqrt{21}< \sqrt{25}=5\)
Vậy \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
cho 3 so thuc x,y,z khac khong va thoa man hai dieu kien \(ax^3=by^3=cz^3\) va \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
chung minh rang : \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
tim tat ca cac so duong a,b,c thoa man dieu kien \(\left\{{}\begin{matrix}\sqrt{a}+\sqrt{b}+\sqrt{c}=6\\\frac{1}{\sqrt{a}}+\frac{4}{\sqrt{b}}+\frac{9}{\sqrt{c}}=6\end{matrix}\right.\)
cho so thuc a,b,c voi a ,b duong va c\(\ne\)0 thoa man
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
1/chung minh c<0 , a+c>0 va b+c >0
2/chung minh \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
1) Cho a,b,c>0 và a+b+c=3
Chứng minh rằng \(\frac{1}{4a^2+b^2+c^2}+\frac{1}{a^2+4b^2+c^2}+\frac{1}{a^2+b^2+4c^2}\le\frac{1}{2}\)
2) Giaỉ phương trình
\(\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}=16-\sqrt{x-2}-\sqrt{y-1}-\sqrt{z-5}\)
cho a,b,c thoa man \(a\times\sqrt{1-b^2}+b\times\sqrt{1-c^2}+c\times\sqrt{1-a^2}=\dfrac{3}{2}\)
chung minh \(a^2+b^2+c^2=\dfrac{3}{2}\)
Cho a, b, c \(\ge\dfrac{-3}{4}\) và a + b + c + d = 3. CMR: \(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\le3\sqrt{7}\)
Cho a, b, c là các số thực dương, thỏa mãn a + b+ c=3. Chứng minh rằng:
\(\frac{a^2}{\sqrt{4a+3b+2}}+\frac{b^2}{\sqrt{4b+3c+2}}+\frac{c^2}{\sqrt{4c+3a+2}}\ge1\)
cho 3 so a,b,c thoa man dieu kien : \(\left\{{}\begin{matrix}a+b+c=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)
tinh gia tri cua bieu thuc T=\(a^2+b^2+c^2\)
Cho a>b>0 .So sánh \(\sqrt{4a+1}-2\sqrt{a}\) và \(\sqrt{4b+1}-2\sqrt{b}\)
