Problem 303 Van Khea

If a, b, x, y, z are positive real numbers such that a+b=1. Prove that
\displaystyle x^3+y^3+z^3+6abxyz\geq (2a-b^2)(x^2y+y^2z+z^2x)+(2b-a^2)(xy^2+yz^2+zx^2)

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Problem 302 Van Khea

If a, b, c are positive real numbers such that a+b+c=3 then prove that
29(a^2b+b^2c+c^2a)+38(ab^2+bc^2+ca^2)+42abc\leq 243

Problem 301 Van Khea

គេអោយ a, b, c ជាចំនួនពិតវិជ្ជមាន។ ស្រាយបញ្ជាក់ថាៈ
Let a, b, c be positive real numbers. Prove that:
\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\geq \frac{3}{2}
Solution
We have
\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\displaystyle =\frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}
Because \displaystyle \frac{3}{2}-\frac{1}{2}=1 then we have
\displaystyle \frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}\displaystyle \geq \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}
Thus, it suffices to show that
\displaystyle \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}\geq \frac{3}{2}
\displaystyle \Leftrightarrow 27((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)\leq 4(a^2+b^2+c^2)^3
Letting a^2=x; b^2=y; c^2=z yields
\displaystyle 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3
From Cauchy-Schwarz inequality we have
\displaystyle ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2\leq (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)
So we need to prove that
\displaystyle (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)\leq \frac{16}{27^2}(x+y+z)^6
Now we will show that for x, y, z are positive real numbers then we have
27(x^2y+y^2z+z^2x+xyz)\leq 4(x+y+z)^3 ;(1) and 27(xy^2+yz^2+zx^2+xyz)\leq 4(x+y+z)^3 ; (2)
without loss of generality, suppose that x=min(x, y, z). Sitting y=x+u and z=x+v; (u, v\geq 0) then:
(1)\Leftrightarrow 9(u^2-uv+v^2)x+(u-2v)^2(4u+v)\geq 0
(2)\Leftrightarrow 9(u^2-uv+v^2)x+(2u-v)^2(u+4v)\geq 0
which is obviously true. Equality occurs for u=v=0\Leftrightarrow x=y=z
\displaystyle \Rightarrow ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2\displaystyle \leq \frac{16}{27^2}(x+y+z)^6
\displaystyle \Leftrightarrow 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3 is true.
Therefore the proof is completed. Equality occurs for x=y=z\Leftrightarrow a=b=c

ទាញយកចំលើយនៅទីនេះ (download here)

VK MaTh 2011: សៀវភៅគណិតសំរាប់សិស្សពូកែ(ជាភាសាវៀតណាម)

Du thi Olympic toan quoc te (2005-2010)

Toan hoc chon loc sinh vien gioi nam 2010

De thi sinh vien gioi cac tinh 2010~2011

Problem 300 Van Khea

If a, b, c are positive real numbers such that a^2+b^2+c^2+2abc=5. Prove that
\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}\geq \frac{4}{3}
Solution
We have
\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}=\frac{1}{a^3+b^3+c^3}+\frac{a^2+b^2+c^2+2abc}{5abc}
\displaystyle \Rightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}+\frac{2}{5}\geq \frac{4}{3}
\displaystyle \Leftrightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}\geq \frac{14}{15}
Let \displaystyle a=\sqrt[3]{\frac{3x}{x+y+z}}; b=\sqrt[3]{\frac{3y}{x+y+z}}; c=\sqrt[3]{\frac{3z}{x+y+z}} then we get a^3+b^3+c^3=3
Sitting a, b, c yields
\displaystyle \frac{1}{3}+\frac{\sqrt[3]{x+y+z}}{5\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{14}{15}
\displaystyle \Leftrightarrow \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq 3
From AM-GM inequality we have
\sqrt[3]{x+y+z}\geq \sqrt[3]{3\sqrt[3]{xyz}} and \displaystyle \sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\geq 3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}
then we get
\displaystyle \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{\sqrt[3]{3\sqrt[3]{xyz}}}{\sqrt[3]{3}}.3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}\geq 3
Therefore the proof is completed. Equality occurs for x=y=z\Leftrightarrow a=b=c=1

Problem 297 Van Khea

If a, b, c are positive real numbers. Prove that:
\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2
Proof
From Cauchy-Schwarz we have
\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(ab+bc+ca)}
But for a, b, c are positive real numbers then 3(ab+bc+ca)\leq (a+b+c)^2
\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{3(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(a+b+c)^2}
Thus, we need to prove that
\displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2
We have \displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\displaystyle =a^2\sqrt{1+\frac{b}{a}}+b^2\sqrt{1+\frac{c}{b}}+c^2\sqrt{1+\frac{a}{c}}\displaystyle =\frac{a^2}{\sqrt{\frac{a}{a+b}}}+\frac{b^2}{\sqrt{\frac{b}{b+c}}}+\frac{c^2}{\sqrt{\frac{c}{c+a}}}\displaystyle \geq \frac{(a+b+c)^2}{\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}}
From Problem Vasile Cirtoaje we have
\displaystyle \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\leq \frac{3}{\sqrt{2}}
\displaystyle \Rightarrow a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2
\displaystyle \Rightarrow (a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2\geq \frac{2}{9}(a+b+c)^4
\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2
Therefore the proof is completed. Equality occurs for a=b=c
របៀបទីពីរ ( Rainymathboy)
សន្មត a\ge b\ge c
ទាញបាន \displaystyle {{a}^{2}}\ge {{b}^{2}}\ge {{c}^{2}}
\displaystyle a+b\ge c+a\ge b+c និង​ \frac{1}{b+c}\ge \frac{1}{c+a}\ge \frac{1}{a+b}
នោះ \displaystyle \frac{{{a}^{2}}}{b+c}\ge \frac{{{b}^{2}}}{c+a}\ge \frac{{{c}^{2}}}{a+b}
តាមវិសមភាព​តំរៀប (Rearrangement​ inequality) យើងមានៈ
បើ​ \displaystyle {{a}_{1}}\ge {{a}_{2}}\ge \ldots \ge {{a}_{n}}​ និង {{b}_{1}}\ge {{b}_{2}}\ge \ldots \ge {{b}_{n}}
តាង \displaystyle A={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+\ldots +{{a}_{n}}{{b}_{n}}
\displaystyle B={{a}_{1}}{{b}_{\sigma (1)}}+{{a}_{2}}{{b}_{\sigma (2)}}+\ldots +{{a}_{n}}{{b}_{\sigma (n)}}
\displaystyle C={{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\ldots +{{a}_{n}}{{b}_{1}}
នាំអោយ A\ge B\ge C
យើងទាញបានៈ
\displaystyle A=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(b+c)
\displaystyle B=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)
\displaystyle C=\frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)
យើងយកលក្ខខណ្ឌ​ B\ge C មក​ប្រើ​នោះយើងទាញបានៈ
\displaystyle \frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)
\displaystyle \ge \frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}
\displaystyle a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល a=b=c

Problem 295 Van Khea (Special problem)

If a, b, c are positive real numbers such that abc=1. Prove that
\displaystyle \frac{a^{\frac{5}{8}}}{\sqrt{b+c}}+\frac{b^{\frac{5}{8}}}{\sqrt{c+a}}+\frac{c^{\frac{5}{8}}}{\sqrt{a+b}}\geq \frac{3}{\sqrt{2}}
Proof
Let a=x^4; b=y^4; c=z^4\Rightarrow xyz=1 then we get:
\displaystyle \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}
From Cauchy-Scharz inequality we have
\displaystyle (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)\displaystyle \geq \biggl(\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}\biggl)^2
Let \displaystyle A=\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}
We have \displaystyle \frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}
Therefore we get \displaystyle A=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}+\frac{(y^3)^{\frac{5}{4}}}{\sqrt[4]{y^4z^4+x^4y^4}}+\frac{(z^3)^{\frac{5}{4}}}{\sqrt[4]{z^4x^4+y^4z^4}}
Because \displaystyle \frac{5}{4}-\frac{1}{4}=1 so from Problem Van Khea inequality we get
\displaystyle A\geq \frac{(x^3+y^3+z^3)^{\frac{5}{4}}}{\sqrt[4]{2(x^4y^4+y^4z^4+z^4x^4)}}\displaystyle \Rightarrow A^2\geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}
\displaystyle \Rightarrow (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)\displaystyle \geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}
\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{(x^3+y^3+z^3)^{\frac{3}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}
Now we will prove that if x, y, z be positive real numbers then
\displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}
Let \displaystyle u=\frac{x\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};v=\frac{y\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};w=\frac{z\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}} then we just need to prove that with u^3+v^3+w^3=3 then (uv)^4+(vw)^4+(wu)^4\leq 3
From AM-GM inequality we have
\displaystyle uv\leq \frac{u^3+v^3+1}{3}=\frac{4-w^3}{3}\displaystyle \Rightarrow (uv)^4\leq \frac{4u^3v^3-u^3v^3w^3}{3}
Therefore we get \displaystyle (uv)^4+(vw)^4+(wu)^4\leq \frac{4((uv)^3+(vw)^3+(wu)^3)}{3}-u^3w^3w^3
Thus, it suffices to show that: 4((uv)^3+(vw)^3+(wu)^3)-3u^3v^3w^3\leq 9
Which is just the third degree Schur's inequality
4(rs+st+tr)(r+s+t)-3rst\leq (r+s+t)^3
For r=u^3; s=v^3; t=w^3
Therefore we get \displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}
\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{3}{2}}}{(x^3+y^3+z^3)^{\frac{4}{3}}\sqrt{2}}\displaystyle \geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{1}{6}}}{\sqrt{2}}
From AM-GM inequality we have x^3+y^3+z^3\geq 3xyz=3
\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}
Therefore the proof is completed. Equality occurs for x=y=z=1\Leftrightarrow a=b=c=1